I'm trying to delete one line from a table, however and even following the examples on stackoverflow, I can't reach a solution.
One of the Important things before start, is that I have the url's hidden, so I don't see the Get url, even the splice don't delete the right line on the table.
HTML:
<button id="{{value.username}}" type='button' type="button" ng-click="delete(value.username, $index)" class="btn btn-primary">Delete</button></td>
JS
$scope.delete = function(deletingId, index){
console.log(deletingId);
$http.get("../admin/deleted.php?username=" + deletingId)
.success(function(data){
$scope.data.splice(index, 1);
console.log('dadasdas');
})
}
PHP
$id = $_GET ['username'];
$sql = "SELECT * FROM members";
$records = mysql_query($sql);
if(isset($_GET['username'])){
$id = $_GET ['username'];
$delete = "DELETE FROM members WHERE username= '$id'";
$res = mysql_query($delete) or die ("FAILED" .mysql_error());
}
Am I doing anything wrong ( the scope is working, the php src is correct, but even doing an echo on php it doesn't fulfill nothing) since I've the url's hidden is there anyway to make it as a post?
Thanks in advance
Try this (I will use Mysqli, because I don't use Mysql);
$servername = "localhost";
$username = "username";
$password = "password";
$dbname = "myDB";
$conn = new mysqli($servername, $username, $password, $dbname);
$id = $_GET['username'];
$sql = "SELECT * FROM users";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
while($row = $result->fetch_assoc()) {
if (strtolower($id) == strtolower($row['username'])) {
$sql = "DELETE FROM users WHERE id=".row['id'];
$conn->query($sql);
}
}
} else {
echo "No accounts registered!";
}
So, deleting stuff with SQL, is case sensitive. So, I hope/think you have a auto increment id in your database. Then this will 100% surely work!
Related
I am creating a database to make my 'PHP' website but I couldn't do this. My website is cruzapp that is related to rideshare companies and changing it in to php to get details about our users. But I can't connect MYSQL by using the following PHP code:
?php
$username = "name";
$password = "password";
$hostname = "host";
//connection to the database
$dbhandle = mysql_connect($hostname, $username, $password)
or die("Not connected to MySQL");
echo "Connected to MySQL<br>";
//select a database to work with
$selected = mysql_select_db("examples",$dbhandle)
or die("Could not select examples");
//execute the SQL query and return records
$result = mysql_query("SELECT id, model,year FROM cars");
//fetch tha data from the database
while ($row = mysql_fetch_array($result)) {
echo "ID:"$row{'id'}." Name:".$row{'model'}."Year: ". //display the results
$row{'year'}.<br>";
}
//close the connection
mysql_close($dbhandle)
?>
Can anyone help me to debug this code?
I will be very thankful to you.
Try this one out. It uses MySQLi with error echoing.
<?php
$username = "name";
$password = "password";
$hostname = "host";
$database = "examples";
$con = mysqli_connect($hostname, $username, $password, $database);
if (!$con) {
exit("Connection failed: " . mysqli_connect_error());
}
$result = mysqli_query($con, "SELECT id, model,year FROM cars");
if (mysqli_error($con)) {
exit("Error: " . mysqli_error($con));
}
while ($row = mysqli_fetch_array($result)) {
echo "ID:" . $row['id'] . " Name:" . $row['model'] . "Year: " . $row['year'] . "<br>";
}
mysqli_close($con);
First of all you should not use mysql because with PHP 7 mysql extension does not work anymore. so you must consider to change it to mysqli or PDO. PDO is recommended. Any how for a quick fix $selected = mysql_select_db($dbhandle,"examples") do this and also check all your values like hostname database name table name and make sure there are no mistakes.
i want to fetch data from my database which is linked to the website,i have created a search box,whenever i enter a particular value in search box,it should display all the related content from the database.i have done the following code,it is not fetching the data from the database,it just shows a blank screen.
<?php
define('db_name','njgh');
define('db_user','root');
define('db_password','');
define('db_host','localhost');
session_start();
$link=mysql_connect(db_host,db_user,db_password);
if(!$link)
{
die('couldnot connetc:'.mysql_error());
}
$db_selected=mysql_select_db(db_name,$link);
if(!$db_selected)
{
die('cant connect to db');
}
function clean($str) {
$str = #trim($str);
if(get_magic_quotes_gpc()) {
$str = stripslashes($str);
}
return mysql_real_escape_string($str);
}
$value = clean($_POST['searchtext']);
$qry = "SELECT * FROM database WHERE Site_ID = '$value'";
$result = mysql_query($qry);
print_r($result);
mysql_query function return resource ID not your data , use mysql_fetch_array, mysql_fetch_object to get data
$value = ($_POST['searchtext']);
$result = mysql_query("SELECT * FROM database WHERE Site_ID = '$value'");
while ($row = mysql_fetch_array($result, MYSQL_NUM)) {
printf("ID: %s Name: %s", $row[0], $row[1]);
}
mysql_* has been removed entirely as of PHP 7.0. Prevent SQL injection and use the mysqli statement class.
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "njgh";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$value = $_POST['searchtext'];
/* create a prepared statement */
if($stmt = $conn->prepare("SELECT * FROM database WHERE Site_ID =?")) {
/* bind parameters for markers */
$stmt->bind_param("s", $value);
/* execute query */
$stmt->execute();
$result = $stmt->get_result();
if($result->num_rows > 0) {
while ($row = $result->fetch_assoc()) {
echo $row['db_table_field_name'];
}
//or
$row = $result->fetch_assoc();
print_r($row);
} else {
}
/* close statement */
$stmt->close();
}
/* close connection */
$con->close();
?>
The original MySQL extension is now deprecated, and will generate E_DEPRECATED errors when connecting to a database. Instead, use the MySQLi or PDO_MySQL extensions.
Hope this will work for you......
<?php
define('db_name','sports');
define('db_user','root');
define('db_password','');
define('db_host','localhost');
$link=mysqli_connect(db_host,db_user,db_password,db_name);
if(!$link)
{
die('couldnot connect:'.mysql_error());
}
function clean($str) {
$link=mysqli_connect(db_host,db_user,db_password,db_name);
$str = #trim($str);
if(get_magic_quotes_gpc())
$str = stripslashes($str);
return mysqli_real_escape_string($link, $str);
}
if(isset($_POST['searchtext'])){
$value = clean($_POST['searchtext']);
$qry = "SELECT * FROM cricket WHERE id = '$value'";
$qrydb = mysqli_query($link, $qry);
while($row=mysqli_fetch_array($qrydb)){
$name = $row['name'];
}
echo $name ;
}
?>
<body>
<form method="post">
<input type="text" name="searchtext">
<input type="submit" name="submit">
</form>
</body>
I am selecting input values from the database and I want to send it to the javascript function addVal() so that I can retrieve this value. I do not want to use echo. It is not working right now and I don't know how I can make it work.
<h1>trial,</h1>
<div id ="val"> </div>
<?php
$name = $_POST['postname'];
$host = 'localhost';
$user = 'root';
$pass = 'root';
$db_name="big";
$conn = new mysqli($host, $user, $pass, $db_name);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
echo "connected";
$sql = "SELECT input FROM trial_db";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
while( $row = $result->fetch_assoc()) {
$value = $row['input'];
addVal ($value);
}
}
?>
<script>
function addVal (value){
document.getElementById("val").innerHTML+= value ;
}
</script>
Calling js function from php will not work remove that code from php.
And change in js.
<script>
function addVal (){
var value = "<?php echo $value; ?>";
document.getElementById("val").innerHTML+= value ;
}
</script>
Will only work if js and php codes are in same php file.
mysql_fetch_array() expects parameter 1 to be resource, boolean given in
<?php
mysql_connect ("localhost", "cab","a321") or die (mysql_error());
mysql_select_db ("ppwxpjey_mcidb");
$termOrd = $_POST['termOrd'];
$sql = mysql_query("select * from booking where order_no like '%$termOrd%'");
while ($row = mysql_fetch_array($sql)){
echo "<table width='1000' border='2' align='center' style='background-color:#FFFFFF;border-collapse:collapse;border:2px solid #6699FF;color:#000000'><tr><th>ORDER NO</th><th>NAME</th><th>MOBILE</th><th>FROM PLACE</th><th>TO PLACE</th><th>JOURNEY DATE</th><th>JOURNEY TIME</th><th>PERSON</th><th>BOOKING TIME</th></tr>";
echo "<tr><td>".$row["ORDER_NO"]."</td><td>".$row["NAME"]."</td><td>".$row["MOBILE"]."</td><td>".$row["FROM_PLACE"]."</td><td>".$row["TO_PLACE"]."</td><td>".$row["JOURNEY_DATE"]."</td><td>".$row["JOURNEY_TIME"]."</td><td>".$row["PERSON"]."</td><td>".$row["UPDATE_TIME"]."</td></tr>";
echo '<br/>';
}
?>
You've to run the query before passing it in mysql_fetch_array which expects resource as a parameter. So change your code like this,
$sql = mysql_query("select * from booking where order_no like '%$termOrd%'");
$result = mysql_query($query) or die(mysql_error());
// This is where you're getting resource or throwing SQL error.
while ($row = mysql_fetch_array($result)){
// YOUR LOGIC.
}
Warning mysql_query, mysql_fetch_array,mysql_connect etc.. extensions were deprecated in PHP 5.5.0, and it was removed
in PHP 7.0.0. Instead, the MySQLi or PDO_MySQL extension should be
used. Source : http://php.net/manual/en/function.mysql-query.php
So update your code as soon as you can.
Use mysqli_* or PDO . mysql_* is deprecated.
The original MySQL extension is now deprecated, and will generate E_DEPRECATED errors when connecting to a database. Instead, use the MYSQLi or PDO_MySQL extensions. use prepared statement
//db connection
global $conn;
$servername = "localhost"; //host name
$username = "cab"; //username
$password = "a321"; //password
$mysql_database = "ppwxpjey_mcidb"; //database name
//mysqli prepared statement
$conn = mysqli_connect($servername, $username, $password) or die("Connection failed: " . mysqli_connect_error());
mysqli_select_db($conn,$mysql_database) or die("Opps some thing went wrong");
$termOrd = "%{$_POST['termOrd']}%";
$stmt = $conn->prepare("select * from booking where order_no like ? ");
$stmt->bind_param('s',$termOrd);
$stmt->execute();
$get_result= $stmt->get_result();
$row_count= $stmt->affected_rows;
if($row_count>0)
{
while($row=$get_result->fetch_assoc())
{
echo "<table width='1000' border='2' align='center' style='background-color:#FFFFFF;border-collapse:collapse;border:2px solid #6699FF;color:#000000'><tr><th>ORDER NO</th><th>NAME</th><th>MOBILE</th><th>FROM PLACE</th><th>TO PLACE</th><th>JOURNEY DATE</th><th>JOURNEY TIME</th><th>PERSON</th><th>BOOKING TIME</th></tr>";
echo "<tr><td>".$row["ORDER_NO"]."</td><td>".$row["NAME"]."</td><td>".$row["MOBILE"]."</td><td>".$row["FROM_PLACE"]."</td><td>".$row["TO_PLACE"]."</td><td>".$row["JOURNEY_DATE"]."</td><td>".$row["JOURNEY_TIME"]."</td><td>".$row["PERSON"]."</td><td>".$row["UPDATE_TIME"]."</td></tr>";
echo '<br/>';
}
}
$stmt->close();
$conn->close();
I can't figure out why jquery ajax post call not working if I used connection on include php file though I'm getting right response from php. Kindly tell me what's the explanation behind this
This is on my DB file
<?php
$host = "mysql2.000webhost.com";
$username = "a212_mt5199";
$password = "secret";
$db = "a211_mydb";
$con = new mysqli ($host, $username, $password, $db);
GLOBAL $con;
if ($con->connect_error) {
die ("Error:" . $con->connect_error);
}
?>
This including the above is not working
<?php
include("db.php");
// set parameters and execute
$myusername=$_POST['name'];
$mypassword=$_POST['pwd'];
// To protect MySQL injection (more detail about MySQL injection)
$myusername = stripslashes($myusername);
$mypassword = stripslashes($mypassword);
$stmt = $con->prepare("SELECT username, password FROM members WHERE username=? and password=?");
$stmt->bind_param("ss", $myusername, $mypassword);
$stmt->execute();
$stmt->bind_result($username, $password);
$stmt->store_result();
// Mysql_num_row is counting table row
$count=$stmt->num_rows;
// If result matched $myusername and $mypassword, table row must be 1 row
if($count==1){
// Register $myusername, $mypassword and redirect to file "login_success.php"
session_register("myusername");
session_register("mypassword");
echo 'true';
$_SESSION['user_name']= $username;
} else {
echo "Wrong Username or Password";
}
$stmt->close();
$con->close();
?>
This is working
<?php
$host = "mysql2.000webhost.com";
$username = "a212_mt5199";
$password = "secret";
$db = "a211_mydb";
$con = new mysqli ($host, $username, $password, $db);
if ($con->connect_error) {
die ("Error:" . $con->connect_error);
}
// set parameters and execute
$myusername=$_POST['name'];
$mypassword=$_POST['pwd'];
// To protect MySQL injection (more detail about MySQL injection)
$myusername = stripslashes($myusername);
$mypassword = stripslashes($mypassword);
$stmt = $con->prepare("SELECT username, password FROM members WHERE username=? and password=?");
$stmt->bind_param("ss", $myusername, $mypassword);
$stmt->execute();
$stmt->bind_result($username, $password);
$stmt->store_result();
// Mysql_num_row is counting table row
$count=$stmt->num_rows;
// If result matched $myusername and $mypassword, table row must be 1 row
if($count==1){
// Register $myusername, $mypassword and redirect to file "login_success.php"
session_register("myusername");
session_register("mypassword");
echo 'true';
$_SESSION['user_name']= $username;
} else {
echo "Wrong Username or Password";
}
$stmt->close();
$con->close();
?>
Able to solve issue by using DEFINE variable.