Ive been trying this for hours now but it wont quite get there.
I have a database which amongst other things contains geocodes, lat and lon. I have accessed these using the following PHP
<?php
mysql_connect("localhost", "tompublic", "public") or die(mysql_error());
mysql_select_db("first_section") or die(mysql_error());
$data = mysql_query("SELECT geo_lat, geo_lon FROM first_page_data")
or die(mysql_error());
while ($row = mysql_fetch_assoc($data)){
$lat[] = $row['geo_lat'];
$lon[] = $row['geo_lon'];
}
?>
These values in $lat and $lon then need to be put into an array in a javascript function like so:
var latit = [];
var longi = [];
latit = '<?php echo $lat[]; ?>';
longi = '<?php echo $lon[]; ?>';
But it wont work! Any ideas?
Try:
var latit = <?php echo json_encode($lat); ?>;
var longi = <?php echo json_encode($lon); ?>;
Edit: Also, the mysql_ functions are deprecated.
You could try this:
var latit = [<?php echo implode(",",$lat) ?>];
var longi = [<?php echo implode(",",$lon) ?>];
First thing is first try to switch to MySQLi due to the fact that Mysql is depreciated.
But try
var latit = <?php echo json_encode($lat); ?>;
var longi = <?php echo json_encode($lon); ?>;
You can either do as #Robbert stated
OR
Using php json_encode convert it to JSON string and you need to JSON.parse() it to convert it to javascript object
var latit = JSON.parse("<?php echo json_encode($lat); ?>");
var longi = JSON.parse("<?php echo json_encode($lon); ?>");
JavaScript arrays are created using a comma separated list surrounded by brackets
var latit = [];
To use your PHP values
var latit = [<?PHP echo implode(",",$lat); ?>];
This assumes your values are numbers. If not, you'll need to include quotes.
var latit = ['<?PHP echo implode("','",$lat); ?>'];
Finally, json_encode is a good option as many of the other answers indicate.
Try this:
<?php
mysql_connect("localhost", "tompublic", "public") or die(mysql_error());
mysql_select_db("first_section") or die(mysql_error());
$data = mysql_query("SELECT geo_lat, geo_lon FROM first_page_data")
or die(mysql_error());
while ($row = mysql_fetch_assoc($data)){
$lat[] = $row['geo_lat'];
$lon[] = $row['geo_lon'];
}
echo '
<script type="text/javascript">
var latit = '.json_encode($lat).';
var longi = '.json_encode($lon).';
</script>
';
?>
Related
basically what I've been trying to do is with PHP get a integer from MySQL after that is done, using JavaScript get the integer that PHP has and display it on HTML
<?php include('ConnectionCode.php');
$conn = mysqli_connect($svr, $usr, $pwd, $db) or die("Could not connect " . mysql_error());
$sql = "SELECT RetailPrice FROM WebHosting_PricingCOP WHERE id='32402' LIMIT 1";
$result = mysqli_query($conn, $sql);
$price = mysqli_fetch_array($result);
echo json_encode(number_format($price['RetailPrice'],0,".",","));
mysqli_close($conn)
?>
the code above will connect to the Database and get the value 59,347
now I'm trying to move this single value from PHP to Javascript in order to display it on HTML
<script type="text/javascript">
var PriceValue = "<?php echo json_encode($price) ?>";
document.write('<h3>'+PriceValue+'</h3>');
</script>
I have gone through many discussions and options here and there and still cant figure out to make it work, when i try to run the html it doesn't display anything
I would greatly appreciate your input
Update:
You may also have a problem with using json_encode inside of "" quotes.
var PriceValue = "<?php echo json_encode($price) ?>";
Instead, use:
var PriceValue = <?php echo json_encode($price) ?>;
or
var PriceValue = <?php echo $price ?>; // if $price is not an object
Note:
To debug this, check the generated HTML source to see what JavaScript you are actually generating.
$price needs to be in the same scope when you try to generate JS.
http://phpfiddle.org/main/code/aeav-gb1w
<?php
$price = 2523525;
?>
<script type="text/javascript">
var PriceValue = "<?php echo $price; ?>";
document.write('<h3>'+PriceValue+'</h3>');
</script>
However, it is going to be preferable for you to use your PHP file as an API endpoint instead of mixing your PHP and JavaScript together.
How to create var javascript via php ?
<?PHP
include("connect.php");
$get_data = mysqli_query($db_mysqli,"SELECT * FROM bad_word");
while($resilt_row = mysqli_fetch_array($get_data))
{
$bad_words = $bad_words."".$resilt_row [word].",";
}
//echo $bad_words;
?>
<script>
var bad_words = ["<?PHP echo $bad_words; ?>"];
alert(bad_words);
</script>
I want to get var javascript like this var bad_words = ["fuck", "ass"];
When alert it's get only blank result.
How can i do that ?
<?php
include("connect.php");
$query = mysqli_query($db_mysqli,"SELECT * FROM bad_word");
$badWords = [];
while($row = mysqli_fetch_array($query))
{
$badWords[] = $row['word'];
}
js
<script>
var bad_words = <?= json_encode($badWords); ?>;
alert(bad_words[0]);
console.log(bad_words);
</script>
I don't like gluing the strings if it's an array then pass it as an array with json_encode() function. Also in views it's better to use short syntax with <?= ?> tag
The string you're making is not the one you want.
You're not adding quotes.
Do this instead, using an array is better:
$get_data = mysqli_query($db_mysqli,"SELECT * FROM bad_word");
$bad_words = array();
while($resilt_row = mysqli_fetch_array($get_data))
{
$bad_words[] = "'" . $resilt_row[word]. "'";
}
And then, down in your js:
var bad_words = ["<?php echo implode(",", $bad_words) ?>"];
or (for short)
var bad_words = ["<?= implode(",", $bad_words) ?>"];
If you want alert them as a string you can do it like this:
<script>
var bad_words = ["<?php echo implode('","',$bad_words); ?>"];
alert(bad_words);
</script>
Otherwise, you can print them as an array:
<script>
var bad_words = <?php echo json_encode($bad_words) ?>;
console.log(bad_words);
</script>
<?PHP
include("connect.php");
$get_data = mysqli_query($db_mysqli,"SELECT * FROM bad_word");
while($resilt_row = mysqli_fetch_array($get_data))
{
$bad_words[] = $resilt_row['word'];
}
//echo $bad_words;
?>
<script>
var bad_words = ["<?php print implode('","',$bad_words); ?>"];
alert(bad_words);
</script>
I am working on a project that requires create hundreds of variables in javascript with PHP values. I can write each one line by line like so:
var M1 = <?php echo json_encode($$mn[0]); ?>;
var M2 = <?php echo json_encode($$mn[1]); ?>;
var M3 = <?php echo json_encode($$mn[2]); ?>;
As I said there are hundreds of these though and if it is possible to do in a loop I would be very interested in learning. I have searched all over and can't find a direct answer. It may very well be that this is not possible. I am new to coding and still learning what certain code can and cannot do.
Any insight or direction on this topic would be greatly appreciated!
If this is not an option is it possible to use an array index for the javascript variable name? I have created an array for the JS and PHP. The PHP works fine above but if I try to use an array index for the JS like below, it breaks:
var mcirc[0] = <?php echo json_encode($$mn[0]); ?>;
I have output the array and the values are coming up correctly but when I run this I get the message:
[object HTMLDivElement]
instead of the actually value that should show up.
UPDATE
$mn array:
for ($m1 = 1; $m1 < 6; $m1++) {
$mn[] = 'M'.$m1;
}
UPDATE
Select SQL creating array:
$sqlMC = "SELECT * FROM tblmaincircles";
$result = $conn->query($sqlMC);
while($row = $result->fetch_assoc()) {
$$row["mcID"]= $row["mcName"];
}
The array for mcID looks like this:
M1 = "text1"
M2 = "text2"
M3 = "text3"
M4 = "text4"
M5 = "text5"
UPDATE
end result desired:
var M1 = "text1";
var M2 = "text2";
var M3 = "text3";
var M4 = "text4";
var M5 = "text5";
Where "text1, ...2, ...3, ...4, ...5" are coming from the MySQL database.
UPDATE
Here is the final code that got this working:
$sqlMC = "SELECT mcID, mcName FROM tblmaincircles";
$result = $conn->query($sqlMC);
while($row = $result->fetch_assoc()) {
$mcID[] = $row["mcID"];
$mcName[] = $row["mcName"];
}
<?php for ($m1 = 0; $m1 <5; $m1++) { ?>
var <?php echo $mcID[$m1]; ?> = <?php echo json_encode($mcName[$m1]); ?>;
<?php } ?>
Simply put JSON into variable
var json = <?php echo json_encode($$mn); ?>;
And then process the JSON way you want:
eg.
var json=[{key:someValue},
{key:someValue2},
{key:someValue3}
];
json.forEach(function(a){
console.log(a.key);
})
First in your query part, declare a variable to hold the result that you want. I'm assuming the M1 is mcID in your table and text1 is the mcName. For example:
$sqlMC = "SELECT * FROM tblmaincircles";
$result = $conn->query($sqlMC);
$mac = [];//or $mac = array(); Depends on your PHP version.
while($row = $result->fetch_assoc()) {
$mac[$row["mcID"]] = $row["mcName"];
}
And then, iterate through the $mac array with foreach loop. I'm assuming you are using PHP codes within HTML. The $key will be the mcID and the $value will be the mcName.
//php tag for the foreach opening
<?php foreach ($mac as $key => $value) { ?>
var <?php echo $key; ?> = <?php echo "'$value';"; ?>
//php tag for the foreach closing
<?php } ?>
OR, if you want to use javascript associative array.
var macJs = {};
<?php foreach ($mac as $key => $value) { ?>
macJs.<?php echo $key; ?> = <?php echo "'$value';"; ?>
<?php } ?>
And you can access the element like this in javascript macJs.M1.
You should use JSON to 'export' your objects/array through different languages, in that case:
var json = '<?= json_encode($your_array); ?>';
After this you can parse this Json, what should return your array:
var your_array = JSON.parse(json);
If I have created a variable in php, say $test, how can I set a variable in javascript, say var test, to be equal to it.
I have already tried var test = <?php $test ?>
I guess
var test = <?php echo json_encode($test) ?>
The naive way var test = '<?php echo ($test) ?>' will fail if $test contains quotes or newlines, let alone is not of the string type (e.g. array, object)
var test = '<?php echo $test; ?>'
try like this :
var test = '<?php echo $test ?>';
var test = '<?php echo $test; ?>';
Or using shorthand echos, like this:
var test = '<?= test;?>';
var test = <?php echo json_encode($test); ?>
You can use
<pre>
var test = '<?php echo $test?>';
</pre>
below the definition of $test.
Change
var test = <?php $test ?>
to
var test = <?php echo $test; ?>
You are missing two things:
var test = <?php $test ?>
1) Echo statement.
2) Single Quotes around PHP snipplet.
So, the corrected code should be:
var test = "<?php echo $test ?>";
Within your page where you want your PHP to output to type:
var MyJavascriptVariable = <?php echo $myPHPVariable; ?>
Advise NOT to use short tags () as this can be disabled by webhosts and may break your code.
I have the following in my page header
<script type="text/javascript" src="https://ajax.googleapis.com/ajax/libs/jquery/1.5.1/jquery.min.js"></script>
<script type="text/javascript" src="javascript/autoSuggest.js"></script>
<script type="text/javascript" src="javascript/suggest.js"></script>
suggest.js is made of:
$(function(){
$("#idName input").autoSuggest("../Test.php", {minChars: 2, matchCase: true});
});
and autoSuggest.js is a plugin by Drew Wilson (http://code.drewwilson.com/entry/autosuggest-jquery-plugin)
Test.php is
<?php
include('database_info.inc');
$input = $_POST["idName"];
$data = array();
var_dump($data);
// query database to see which entries match the input
$query = mysql_query("SELECT * FROM test WHERE title LIKE '%$input%'");
while ($row = mysql_fetch_assoc($query)) {
$json = array();
$json['value'] = $row['id'];
$json['name'] = $row['title'];
$data[] = $json;
}
header("Content-type: application/json");
echo json_encode($data);
?>
My var_dump() doesn't do anything and no items are suggested ...what could I be doing wrong? seems like there's no communication with Test.php
A quick look at the plug-in suggests it expects a parameter called q passed as a GET string, not as a POST.
<?
$input = $_GET["q"];
$data = array();
// query your DataBase here looking for a match to $input
$query = mysql_query("SELECT * FROM my_table WHERE my_field LIKE '%$input%'");
while ($row = mysql_fetch_assoc($query)) {
$json = array();
$json['value'] = $row['id'];
$json['name'] = $row['username'];
$json['image'] = $row['user_photo'];
$data[] = $json;
}
header("Content-type: application/json");
echo json_encode($data);
?>