I want to take two strings and returns the longer string. If both strings have the same length, then the function should return the string 'TIE'. I am very new to javascript
function getLongerString(str1,str2)
{
var a = console.log("str1");
var b = console.log("str2");
if(a.length==b.length)
{
long = "TIE";
}
else if(a.length>b.length)
{
long = a;
}
else
{
long = b;
}
console.log(long);
}
The issue is with this two lines
var a = console.log("str1").value;
var b = console.log("str2").value;
Also no need to create intermediate variable a & b
It does not make anything logical
function getLongerString(str1, str2) {
let long = "";
if (str1.length === str2.length) {
long = "TIE";
} else if (str1.length > str2.length) {
long = str1;
} else {
long = str2;
}
console.log(long);
}
getLongerString("Hello World", "New World")
Check this out:
var pre = onload, getLongerStr; // for use on other loads
onload = function(){
if(pre)pre(); // change var name if using technique on another page
function getLongerStr(str1, str2){
var s1 = str1.length, s2 = str2.length;
if(s1 === s2){
return 'TIE';
}
else{
return s1 > s2 ? str1 : str2;
}
}
console.log(getLongerStr('what', 'cool'));
console.log(getLongerStr('lame', 'yes'));
console.log(getLongerStr('absolutely', 'pointless'));
}
Related
I have built this function to replace a group of characters in a string by a random value from another list within a function:
function replaceExpr(a) {
var expToReplace = 0
var newSent = a
while (expToReplace == 0) {
if (a.search("zx") == -1) {
expToReplace = 1
} else {
var startPos = a.search("zx");
startPos += 2;
var endPos = a.search("xz");
var b = a.substring(startPos, endPos);
var fn = window[b];
if (typeof fn === "function") var newWord = fn();
final = newSent.replace("zx" + b + "xz", newWord);
newSent = final
a = a.replace("zx" + b + "xz", "")
}
}
return final
}
function appearance() {
var list = [
"attractive",
"fit",
"handsome",
"plain",
"short",
"tall",
"skinny",
"well-built",
"unkempt",
"unattractive"
]
return list[Math.floor(Math.random() * list.length)];
}
function personality() {
var list = [
"aggresive",
"absent-minded",
"cautious",
"detached from the real world",
"easygoing",
"focused",
"honest",
"dishonest",
"polite",
"uncivilized"
]
return list[Math.floor(Math.random() * list.length)];
}
An example :
var a = replaceExpr("Theodor is a zxappearancexz man. He seems rather zxpersonalityxz.")
alert(a)
// Theodor is a unattractive man. He seems rather cautious.
Everything works perfectly with the function but I have an issue related to it. As you can see, there's one grammar mistake : it's written "a unattractive" where it should be "an unattractive".
There's a function I usually use to to fix the a\an issue which is :
var AvsAnSimple = (function (root) {
//by Eamon Nerbonne (from http://home.nerbonne.org/A-vs-An), Apache 2.0 license
// finds if a word needs a "a" or "an" before it
var dict = "2h.#2.a;i;&1.N;*4.a;e;i;o;/9.a;e;h1.o.i;l1./;n1.o.o;r1.e.s1./;01.8;12.1a;01.0;12.8;9;2.31.7;4.5.6.7.8.9.8a;0a.0;1;2;3;4;5;6;7;8;9;11; .22; .–.31; .42; .–.55; .,.h.k.m.62; .k.72; .–.82; .,.92; .–.8;<2.m1.d;o;=1.=1.E;#;A6;A1;A1.S;i1;r1;o.m1;a1;r1; .n1;d1;a1;l1;u1;c1.i1.a1.n;s1;t1;u1;r1;i1;a1;s.t1;h1;l1;e1;t1;e1.s;B2.h2.a1.i1;r1;a.á;o1.r1.d1. ;C3.a1.i1.s1.s.h4.a2.i1.s1;e.o1.i;l1.á;r1.o1.í;u2.i;r1.r1.a;o1.n1.g1.j;D7.a1.o1.q;i2.n1.a1.s;o1.t;u1.a1.l1.c;á1. ;ò;ù;ư;E7;U1;R.b1;o1;l1;i.m1;p1;e1;z.n1;a1;m.s1;p5.a1.c;e;h;o;r;u1.l1;o.w1;i.F11. ;,;.;/;0;1;2;3;4;5;6;71.0.8;9;Ae;B.C.D.F.I2.L.R.K.L.M.N.P.Q.R.S.T.B;C1;M.D;E2.C;I;F1;r.H;I3.A1;T.R1. ;U;J;L3.C;N;P;M;O1. ;P1;..R2.A1. ;S;S;T1;S.U2.,;.;X;Y1;V.c;f1.o.h;σ;G7.e1.r1.n1.e;h1.a3.e;i;o;i1.a1.n1.g;o2.f1. ;t1.t1. ;r1.i1.a;w1.a1.r1.r;ú;Hs. ;&;,;.2;A.I.1;2;3;5;7;B1;P.C;D;F;G;H1;I.I6;C.G.N.P.S1.D;T.K1.9;L;M1;..N;O2. ;V;P;R1;T.S1.F.T;V;e2.i1.r;r1.r1.n;o2.n6;d.e1.s;g.k.o2;l.r1;i1.f;v.u1.r;I3;I2;*.I.n1;d1;e1;p1;e1;n1;d2;e1;n1;c1;i.ê.s1;l1;a1;n1;d1;s.J1.i1.a1.o;Ly. ;,;.;1;2;3;4;8;A3. ;P;X;B;C;D;E2. ;D;F1;T.G;H1.D.I1.R;L;M;N;P;R;S1;m.T;U1. ;V1;C.W1.T;Z;^;a1.o1.i1.g;o1.c1.h1.a1;b.p;u1.s1.h1;o.ộ;M15. ;&;,;.1;A1;.1;S./;1;2;3;4;5;6;7;8;Ai;B.C.D.F.G.J.L.M.N.P.R.S.T.V.W.X.Y.Z.B1;S1;T.C;D;E3.P1;S.W;n;F;G;H;I4. ;5;6;T1;M.K;L;M;N;O1.U;P;Q;R;S;T1;R.U2. ;V;V;X;b1.u1.m;f;h;o2.D1.e.U1;..p1.3;s1.c;Ny. ;+;.1.E.4;7;8;:;A3.A1;F.I;S1.L;B;C;D;E3.A;H;S1. ;F1;U.G;H;I7.C.D1. ;K.L.N.O.S.K;L;M1;M.N2.R;T;P1.O1.V1./1.B;R2;J.T.S1;W.T1;L1.D.U1.S;V;W2.A;O1.H;X;Y3.C1.L;P;U;a1.s1.a1.n;t1.h;v;²;×;O5;N1;E.l1;v.n2;c1.e.e1.i;o1;p.u1;i.P1.h2.i1.a;o2.b2;i.o.i;Q1.i1.n1.g1.x;Rz. ;&;,;.1;J./;1;4;6;A3. ;.;F1;T.B1;R.C;D;E3. ;S1.P;U;F;G;H1.S;I2.A;C1. ;J;K;L1;P.M5;1.2.3.5.6.N;O2.H;T2;A.O.P;Q;R1;F.S4;,...?.T.T;U4;B.M.N.S.V;X;c;f1;M1...h2.A;B;ò;S11. ;&;,;.4.E;M;O;T1..3.B;D;M;1;3;4;5;6;8;9;A3. ;8;S2;E.I.B;C3.A1. ;R2.A.U.T;D;E6. ;5;C3;A.O.R.I1.F.O;U;F3;&.H.O1.S.G1;D.H3.2;3;L;I2. ;S1.O.K2.I.Y.L3;A2. ;.;I1. ;O.M3;A1. ;I.U1.R.N5.A.C3.A.B.C.E.F.O.O5. ;A1.I;E;S1;U.V;P7;A7;A.C.D.M.N.R.S.E1. ;I4;C.D.N.R.L1;O.O.U.Y.Q1. ;R;S1;W.T9.A1. ;C;D;F;I;L;M;S;V;U7.B.L.M.N.P.R.S.V;W1.R;X1.M;h1.i1.g1.a1.o;p1.i1.o1;n.t2.B;i1.c1.i;T4.a2.i2.g1.a.s1.c;v1.e1.s;e1.a1.m1.p;u1.i2.l;r;à;Um..1.N1..1.C;/1.1;11. .21.1;L1.T;M1.N;N4.C1.L;D2. .P.K;R1. .a;b2;a.i.d;g1.l;i1.g.l2;i.y.m;no. ;a1.n.b;c;d;e1;s.f;g;h;i2.d;n;j;k;l;m;n;o;p;q;r;s;t;u;v;w;p;r3;a.e.u1.k;s3. ;h;t1;r.t4.h;n;r;t;x;z;í;W2.P1.:4.A1.F;I2.B;N1.H.O1.V;R1.F1.C2.N.U.i1.k1.i1.E1.l1.i;X7;a.e.h.i.o.u.y.Y3.e1.t1.h;p;s;[5.A;E;I;a;e;_2._1.i;e;`3.a;e;i;a7; .m1;a1;r1. .n1;d2; .ě.p1;r1;t.r1;t1;í.u1;s1;s1;i1. .v1;u1;t.d3.a1.s1. ;e2.m1. ;r1. ;i2.c1.h1. ;e1.s1.e2.m;r;e8;c1;o1;n1;o1;m1;i1;a.e1;w.l1;i1;t1;e1;i.m1;p1;e1;z.n1;t1;e1;n1;d.s2;a1. .t4;a1; .e1; .i1;m1;a1;r.r1;u1.t.u1.p1. ;w.f3. ;M;y1.i;h9. ;,;.;C;a1.u1.t1;b.e2.i1.r1;a.r1.m1.a1.n;o4.m2.a1; .m;n8; .b.d.e3; .d.y.g.i.k.v.r1.s1. ;u1.r;r1. ;t1;t1;p1;:.i6;b1;n.e1;r.n2;f2;l1;u1;ê.o1;a.s1;t1;a1;l1;a.r1; .s1; .u.k1.u1. ;l3.c1.d;s1. ;v1.a;ma. ;,;R;b1.a.e1.i1.n;f;p;t1.a.u1.l1.t1.i1.c1.a1.m1.p1.i;×;n6. ;V;W;d1; .t;×;o8;c2;h1;o.u1;p.d1;d1;y.f1; .g1;g1;i.no. ;';,;/;a;b;c1.o;d;e2.i;r;f;g;i;l;m;n;o;r;s;t;u;w;y;z;–;r1;i1;g1;e.t1;r1.s;u1;i.r3. ;&;f;s9.,;?;R;f2.e.o.i1.c1.h;l1. ;p2.3;i1. ;r1.g;v3.a.e.i.t2.A;S;uc; ...b2.e;l;f.k2.a;i;m1;a1. .n3;a3; .n5.a;c;n;s;t;r1;y.e2; .i.i8.c2.o1.r1.p;u1.m;d1;i1.o;g1.n;l1.l;m1;o.n;s1.s;v1.o1;c.r5;a.e.i.l.o.s3. ;h;u1.r2;e.p3;a.e.i.t2.m;t;v.w1.a;xb. ;';,;.;8;b;k;l;m1;a.t;y1. ;y1.l;{1.a;|1.a;£1.8;À;Á;Ä;Å;Æ;É;Ò;Ó;Ö;Ü;à;á;æ;è;é1;t3.a;o;u;í;ö;ü1; .Ā;ā;ī;İ;Ō;ō;œ;Ω;α;ε;ω;ϵ;е;–2.e;i;ℓ;";
function fill(node) {
var kidCount = parseInt(dict, 36) || 0,
offset = kidCount && kidCount.toString(36).length;
node.article = dict[offset] == "." ? "a" : "an";
dict = dict.substr(1 + offset);
for (var i = 0; i < kidCount; i++) {
var kid = node[dict[0]] = {}
dict = dict.substr(1);
fill(kid);
}
}
fill(root);
return {
raw: root,
//Usage example: AvsAnSimple.query("example")
//example returns: "an"
query: function (word) {
var node = root, sI = 0, result, c;
do {
c = word[sI++];
} while ('"‘’“”$\''.indexOf(c) >= 0);//also terminates on end-of-string "undefined".
while (1) {
result = node.article || result;
node = node[c];
if (!node) return result;
c = word[sI++] || " ";
}
}
};
})({})
Now, the problem is that I can't find a way to use this function in conjunction with the replaceExpr. The following obviously wouldn't work because of order precedence :
var a = replaceExpr("Theodor is " + AvsAnSimple(zxappearancexz) + "man. He seems rather " + AvsAnSimple(zxpersonalityxz).")
I just recently started learning javascript so my knowledge is rather limited. Any ideas how I could overcome this?
Thank you!
You could use a regular expression to optionally match the " a " or "an" before your word in the input string and store that matched portion in a variable using the String.match() function, then check if that " a " or " an " exists in your matched string, do the manipulations you need to do and store that manipulated string in a separate variable, then use String.replace() to find that previously matched string again, and replace it wit
your manipulated string. The regular expression you could use for this is /(\san?\s)?(zx\w*zx)/gm
See the regular expression here for more context.
Thank you Joseph! With your help I managed to find something that works by using your regular expression. Here's my function :
function replaceExpr(a) {
var nbExprToReplace = 1;
while (nbExprToReplace == 1) {
if (a.search("zx") == -1) {
nbExprToReplace = 0;
} else {
var currentGroup = a.match(/(\san?\s)?(zx\w*xz)/);
var exprToChange = currentGroup[2];
exprToChange = exprToChange.slice(2,-2);
var exprToChange = window[exprToChange];
if (typeof exprToChange !== "function") {
alert("the keyword is not a recognized function!");
break;
} else {
exprToChange = exprToChange();
var final = exprToChange
};
if (currentGroup[1] === undefined) {
} else {
var newArticle = AvsAnSimple.query(exprToChange);
final = newArticle.concat(" " + final)
};
a = a.replace(currentGroup[0], " " + final);
};
};
return a;
};
I am writing code for recursion. And here is my code.
Here, what I am trying to do is, if string has ' then replace it with HTML quotes and calling function recursively until all ' have been replaced.
But this is always returning me false. When I alert var a. If I not use return false then it returns undefined. Any clue what is the wrong here?
var a = replaceqt(" hello's there 'how are you?' ");
console.log(a);
function replaceqt(object) {
var indexc = object.indexOf("'");
var next = object.charAt(indexc + 1);
var prev = object.charAt(indexc - 1);
if (indexc == 0) {
object = object.replace("'", "‘");
} else if (parseInt(prev) >= parseInt(0) && parseInt(prev) <= parseInt(9)) {
object = object.replace("'", "'");
} else if (next == " ") {
object = object.replace("'", "’");
} else if (prev == " ") {
object = object.replace("'", "‘");
} else {
object = object.replace("'", "’");
}
indexc = object.indexOf("'");
if (indexc > -1) {
replaceqt(object);
return false;
} else {
return object;
}
}
Because you are returning false whenever there is a second call. Should return the result of recursive invocation instead.
var a = replaceqt(" hello's there 'how are you?' ");
console.log(a);
function replaceqt(object) {
var indexc = object.indexOf("'");
var next = object.charAt(indexc + 1);
var prev = object.charAt(indexc - 1);
if (indexc == 0) {
object = object.replace("'", "‘");
} else if (parseInt(prev) >= parseInt(0) && parseInt(prev) <= parseInt(9)) {
object = object.replace("'", "'");
} else if (next == " ") {
object = object.replace("'", "’");
} else if (prev == " ") {
object = object.replace("'", "‘");
} else {
object = object.replace("'", "’");
}
indexc = object.indexOf("'");
if (indexc <= -1) {
return object;
}
return replaceqt(object);
}
BTW you don't need parseInt(num) if num is a number say 0 or 9.
You need to replace
if (indexc <= -1){
return object;
}else{
replaceqt(object); return false;
}
with
if (indexc <= -1){
return object;
}else{
return replaceqt(object);
}
In your original code, the return value of replaceqt(object) is discarded when indexc >= 0.
You should try using .split and .join functions to simplify your code.
For a simple find-replace all, you can do this:
var sentence = "I hate spaces."
var charToFind = " ";
var replacement = "-";
var afterSplit = sentence.split(charToFind) // ["I", "hate", "spaces"]
var result = afterSplit.join(replacement) // "I-hate-spaces"
Your example is more complex than a find replace, because you need to keep track of left and right quotes.
To get around that, we can figure out if it's even or odd using the index in the array.
var someString = "My 'name' is 'Ryan'... I 'think'."
function replaceQuotesFor (str) {
return str
.split("'")
.map(function (str, index) {
var quote = index % 2 === 1
? '‘'
: '’'
return (index === 0)
? str
: quote + str
})
.join('')
}
console.log('Before:', someString)
console.log('After:', replaceQuotesFor(someString))
I stopped using for loops and modifying indices, because it made debugging frustrating.
I hope these functions help simplify your code and help you in the future!
I am trying to figure out a javascript function that will help resolve this test. I need to be able to determine if the string of words (var matches) that is given is an anagram of the word that I am running through (var subject). In this case there would not be a match. Any and all help will be greatly appreciated. Thank you in advance!
var anagram = require('./anagram');
describe('Anagram', function() {
it("no matches",function() {
var subject = anagram("diaper");
var matches = subject.matches([ "hello", "world", "zombies", "pants"]);
expect(matches).toEqual([]);
});
});
This is what I have so far:
for (var i = 0; i < matches.length; i++) {
if (subject.length != matches[i].length) {
return false
} else if (subject.length == matches[i].length){
var anagram = function(subject, matches) {
return subject.split("").sort("").join("") === matches[i].split("").sort("").join("");
};
}
Here is the fiddle:
http://jsfiddle.net/hn8r4v3u/2/
I alphabetized the letters within the word, as you were doing, in a function.
function getAlphaSortedWord(word) {
var baseWordCharArray = word.split("");
baseWordCharArray.sort();
return baseWordCharArray.join("");
}
The code has a set up:
var baseWord = getAlphaSortedWord("bob");
var thingsToCheck = ["obb", "2", "bob", "", "bo", "ob"];
And then solves it two ways, once with filter and once without it.
var matches = _.filter(thingsToCheck, function (str) {
return (baseWord === getAlphaSortedWord(str));
});
var matches2 = [];
for (index = 0; index < thingsToCheck.length; index++) {
if (baseWord === getAlphaSortedWord(thingsToCheck[index])) {
matches2.push(thingsToCheck[index]);
}
}
You should be able to use these to tie in with your real data for the test to pass.
NOTE, I would add some sanity for "is string" to my function if this is going to be production code.
Found here and it works: https://gist.github.com/AlbertoElias/10005056
function areAnagrams(a, b) {
var c = false;
if (a.length !== b.length) {
return c;
}
var hashMap = {};
var char;
var i;
for (i=0;i<a.length;i++) {
char = a[i];
hashMap[char] = hashMap[char] !== undefined ? hashMap[char]+1 : 1;
}
for (i=0;i<b.length;i++) {
char = b[i];
if (hashMap[char] !== undefined) {
if (hashMap[char] > 1) {
hashMap[char]--;
} else {
delete hashMap[char];
}
} else {
return c;
}
}
if (Object.keys(hashMap).length === 0) c = true;
return c;
}
var fNum = parseFloat("32.23.45"); results in 32.23 but I need the string from last decimal point: 23.45
For example, the following strings should return the following values:
"12.234.43.234" -> 43.234,
"345.234.32.34" -> 32.34 and
"234.34.34.234w" -> 34.34
A fairly direct solution:
function toFloat(s) {
return parseFloat(s.match(/\d+(\.|$)/g).slice(-2).join('.'));
}
For example:
toFloat("32.23.45") // 23.45
toFloat("12.234.43.234") // 43.234
toFloat("345.234.32.34") // 32.34
toFloat("234.34.34.234w") // 34.34
Update: Here's an alternative version which will more effectively handle strings with non-digits mixed in.
function toFloat(s) {
return parseFloat(s.match(/.*(\.|^)(\d+\.\d+)(\.|$)/)[2]);
}
The following will do exactly what you would like (I'm presuming that the last one should return 34.234, not 34.24).
alert (convText("12.234.43.234"));
alert (convText("345.234.32.34"));
alert (convText("234.34.34.234w"));
function convText(text) {
var offset = text.length - 1;
var finished = false;
var result = '';
var nbrDecimals = 0;
while(!finished && offset > -1) {
if(!isNaN(text[offset]) || text[offset] === '.') {
if(text[offset] === '.') {
nbrDecimals++;
}
if(nbrDecimals > 1) {
finished = true;
} else {
result = text[offset] + result;
}
}
offset--;
}
return result;
}
This is my code:
var Evalcard = function(number) {
if (number == 1) {
this.name = "Ace";
this.value = 11;
}
else if (number == 11) {
this.name = "Jack";
this.value = 10;
}
else if (number == 12) {
this.name = "Queen";
this.value = 10;
}
else if (number == 13) {
this.name = "King";
this.value = 10;
}
return {this.name,this.value};
I'm pretty sure this return statement is not correct. How do you make a function return more than one value? Any help at all would be great.
In this case, you probably want to return either an array or an object literal:
return { name: this.name, value: this.value };
// later: EvalCard(...).name; EvalCard(...).number;
return [ this.name, this.value ];
// later: EvalCard(...)[0]; EvalCard(...)[1];
How about this:
return [this.name, this.value];
You could pass an object literal as you came so close to doing:
return { name:this.name, value:this.value };
or you could pass an array:
return [this.name, this.value];
Of course if your code is executed in the global context, you'll be setting name and value on the window object. If you're using Evalcard as a constructor, you wont need a return statement, the object being created will automatically be set:
var e = new Evalcard(1);
console.log(e.name); //outputs "Ace" if you remove the return statement.
Try:
return [this.name, this.value];
Try this...
function xyz() {
...
var x = 1;
var y = 'A';
return [x, y];
}
var a = xyz();
document.write('x=' + a[0] + ' and y = ' + a[1]);
Working example: http://jsfiddle.net/CxTWt/
var Evalcard = function(number) {
var evalName, evalValue;
if (number == 1) {
evalName= "Ace";
evalValue = 11;
}else if (number == 11) {
evalName = "Jack";
evalValue = 10;
}else if (number == 12) {
evalName= "Queen";
evalValue= 10;
}else if (number == 13) {
evalName= "King";
evalValue = 10;
}
return {name: evalName, value: evalValue};
}
alert(Evalcard(1).name+" "+Evalcard(1).value);
You need to change it to return an array or give keys to the object you are returning
So
return [this.name,this.value];
Or
return {name:this.name,value:this.value};
I would return an object:
return {key1:value1, key2:value2}
Then you can reference it like so:
myReturn.key1;
You can return it in a number of different ways:
Array
return [this.name,this.value];
Object
return {first:this.name, second:this.value};
String
return this.name+":"+this.value;