Mysql data retrieve in PHP - php

Question

I'm trying to execute a simple mySQL query in a php page, and I keep getting this error :
"Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in..."
even though the query returns a result in mysql workbench.
This is my code:
<?php
$con=mysqli_connect("localhost","root","","eshkol");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$sql1="SET Names utf8";
$sql = mysql_query("SELECT * FROM student WHERE idStudent=2");
$r = mysql_fetch_array($sql);
echo $r["idStudent"];
if (!mysqli_query($con,$sql1))
{
die('Error hebrew: ' . mysqli_error($con));
}
if (!mysqli_query($con,$sql))
{
die('Error: ' . mysqli_error($con));
}
echo "success";
mysqli_close($con);
?>
What am I doing wrong here?

Answer 1

You're mixing mysql_* and mysqli_* functions.
$sql = mysql_query($con, "SELECT * FROM student WHERE idStudent=2");
$r = mysql_fetch_array($sql);
should be
$sql = mysqli_query($con, "SELECT * FROM student WHERE idStudent=2");
$r = mysqli_fetch_array($sql);
What's interesting is you're using them just below that code:
if (!mysqli_query($con,$sql1))
{
die('Error hebrew: ' . mysqli_error($con));
}
if (!mysqli_query($con,$sql))
{
die('Error: ' . mysqli_error($con));
}
You probably want to combine the two to clean up your code:
$sql = mysql_query($con, "SELECT * FROM student WHERE idStudent=2");
if (!$sql) {
die('Error: ' . mysqli_error($con));
}
$r = mysql_fetch_array($sql);