Use of PHP script in a jquery function

Question

I've echoed the following php and it shows up properly in HTML so that can't be a problem:

PHP

if (mysqli_num_rows($result) > 0) {

//create the drop down menu
 $list ="";
 while($row = mysqli_fetch_assoc($result)) {
     $list = '<div class="dropOption">'.$row["item"].'</div>';
 }

This outputs three rows - apple, pear, strawberry - in the right div format.

And when I put the following php script into the jquery function below, the menu does contain the value strawberry (the last), however the first two are missing.

JavaScript

//drop down menu
$(document).ready(function(){
function createDropdown(){
    var drop = $('#customDropdown');
    var i;
    var htmlString = '<div id="dropContainer">';

    htmlString += '<?php echo $list;?>';

    htmlString += '</div>';
    drop.append(htmlString);
    }

   createDropdown();

I'm new to jquery and php so forgive me if the error is simple; however I'm pretty sure it's right, functionally speaking, because I'm getting something; so I figured the syntax must get be wrong somewhere. Can anybody help? Thanks in advance.

curveball · Accepted Answer · 2017-02-09 13:46:37Z

3

In your php it should be

$list .= '<div class="dropOption">'.$row["item"].'</div>';

and then output just like this <?=$list;?> in your js part

. is used to concatenate strings in php

edited Feb 9, 2017 at 13:46

answered Feb 9, 2017 at 13:40

curveball

4,51315 gold badges47 silver badges53 bronze badges

Sign up to request clarification or add additional context in comments.

3 Comments

user1849962 Over a year ago

like this you mean - $list = '<div class="dropOption">".$row["item"]."</div>'; if so, it didn't seem to work

curveball Over a year ago

I mean leave it as it was in your JS part and add '.' in your php part as I wrote in my answer.

user1849962 Over a year ago

Please note, the echo part is still needed in the output, before the variable list.

Tama · Accepted Answer · 2017-02-09 13:44:29Z

1

You're only assigning the last item to the $list variable. You need to concatenate them all together.

Try:

if (mysqli_num_rows($result) > 0) {
    // Create the drop down menu
    $list = "";
    while($row = mysqli_fetch_assoc($result)) {
        $list .= '<div class="dropOption">' . $row["item"] . '</div>';
    }
}

Note the $list .= instead of $list = part.

answered Feb 9, 2017 at 13:44

Tama

1,7451 gold badge15 silver badges11 bronze badges

Collectives™ on Stack Overflow

Use of PHP script in a jquery function

2 Answers 2

3 Comments

Comments

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

3 Comments

Comments

Related