Pass php string as parameter for javascript function called in php script

Question

So I have this serie of radio buttons generated by a php script that call a javascript function when they get clicked on:

<?php
    $databases = mysqli_query($link, "SHOW DATABASES");
    while($row = mysqli_fetch_row($databases)) {
        $db = $row[0];
        echo "<input type='radio' name='database' id='".$db."' onclick='displaydbcontent(".$db.")'>";     
    }
?>

The javascript function then has to alert the content of the string:

function displaydbcontent(dbid) {
    alert("This is a test");
    alert(dbid);
}

Now, no matter what the content of the variable $db is, after displaying the first alert printing "This is a test", the second one always prints "[object HTMLInputElement]".

I am pretty sure that it's all a matter of quotation marks, but I don't see any way to work around it.

And for those who are wondering about it even thought it hasn't to do much with my issue, the $link variable has been initializated and works correctly.

Adding same id and name for different html elements in the page is not a good sense.. — Mohammedshafeek C S
– Mohammedshafeek C S, Commented Apr 16, 2016 at 12:54
Because you're passing an object to the function, not a string. — dmeglio
– dmeglio, Commented Apr 16, 2016 at 13:45

Rayon · Accepted Answer · 2016-04-18 04:19:55Z

1

Pass this.id as argument which will return id property of the clicked-element

onclick='displaydbcontent(this.id)'>";

As you are passing $db as argument, it represents the element. Element with id becomes the global variables.

function callMe(elem) {
  console.log(elem);
}

<script src="http://gh-canon.github.io/stack-snippet-console/console.min.js"></script>
<div id='elem' onclick='callMe(elem)'>Hello!!!!!</div>

Edit: Or echo '<input type="radio" name="database" id="'.$db.'" onclick="displaydbcontent(\'$db\')">';

edited Apr 18, 2016 at 4:19

answered Apr 16, 2016 at 12:45

Rayon

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2 Comments

Rayon Over a year ago

Happy to help! Kindly accept and up-vote the best solution which has solved the purpose :) Happy coding!

Alex Over a year ago

Just to clarify, the this.id solution works, the other one doesn't since javascript quotation marks mess up with php's.

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