Concatening Php Variable And Strings Inside Double Quotes For Onclick Event

The button when pressed call a function that needs the sn value, but the code below fails with the Chrome debug message: Uncaught ReferenceError: Telephony is not defined Telepho

Solution 1:

1. Assume your variable $sn is "Stack". Hence $sn is a string you need to pass this variable as a string.
2. So rewrite myF($sn) as myF("'".$sn."'").
3. When yu inspect the button you can see myF('Stack').
4. then in javascript you can avoid the error.

Solution 2:

Why do you need double quotes?

echo"<button onclick=\"myF($sn)\">Vote</button>";

Do this instead:

echo'<button onclick="myF(\''. $sn . '\')">Vote</button>';

Or perhaps do it with sprintf like this:

echo sprintf("<button onclick=\"myF('%s')\">Vote</button>", $sn);

Solution 3:

Does this work?

echo"<button onclick=\"myF('$sn')\">Vote</button>";

I tried out a simple test case, without any MySQL. You can find the code I used below:

<?php$sn='noname';
    $registers = array(array('sname' => 'apple'), array('sname' => 'banana'), array('sname' => 'woodapple')); // Dummy "MySQL" Fetch Array (this might differ from what your 'while' loop receives)foreach ($registersas$reg){ // In your case, a 'while' loop works.echo"Service Name: ".$reg['sname']."<br>";
     $sn = $reg['sname'];
     echo"<button onclick=\"myF('$sn')\">Vote</button>"; // Added Quotes.echo"<hr>";
    }
?>

HINT: Please refrain from using mysql_* functions! They are deprecated!

Solution 4:

Your concatenation need to change like this,

echo'<button onclick="myF("'.$sn.'")">Vote</button>';

Solution 5:

I think that when you are passing the PHP variable as an argument in to the onclick event of button for calling javascript function, you string breaks because it is not quoted. So it needs to be quoted. I have edited the code line where you need to edit you code. So try like:

...
......
........
........
echo"<button onclick=\"myF(&quot;$sn&quot;)\">Vote</button>";
.......
......
....