CODEWITHSUNDEEP
javascriptphpjqueryhtml
jak7890
jak7890

Reputation: 11

jquery .show() by php variable

I'm trying to trigger jquery .show() events triggered by php variables.

Here's my .php (which draws from a form submission on a separate page):

    $last_pic_displayed = trim($_POST["last_pic_displayed"]);
    if ( strlen($last_pic_displayed) <= 0 )
        {  $last_pic_displayed = trim($_GET["last_pic_displayed"]);  }

and here's my .show() code:

    <script>
    var lpic = "<?php echo $last_pic_displayed; ?>";
    lpic = parseInt(lpic);
    if (lpic == 0) {
        $("#1").show();
    }
    </script>

I use variations on this script for multiple divs with the intention that the correct $last_pic_displayed; value will trigger the correct div to show, but so far none have.

Here's an example of the form that's posting data to my php:

    <form id="buttons" action="process_form.php" method="post">

        <label for="child_name"><img src="images/enter_name_text.png" alt="Enter Your Name" id="name_text"></label>
        <input type="text" id="child_name" name="child_name" />



            <section>
                <input checked="checked" id="radio1" name="last_pic_displayed" type="radio" value="0" />
                <label for="radio1"><img src="images/3_letter_text.png" alt="3 Letter Words" class="level_text">
                </label>
            </section>

            <section>
                <input checked="checked" id="radio2" name="last_pic_displayed" type="radio" value="6" />
                <label for="radio1"><img src="images/4_letter_text.png" alt="4 Letter Words" class="level_text">
                </label>
            </section>

            <section>
                <input checked="checked" id="radio3" name="last_pic_displayed" type="radio" value="13" />
                <label for="radio1"><img src="images/5_letter_text.png" alt="5 Letter Words" class="level_text">
                </label>
            </section>

            <section id="submit_button">
                <input type="image" name="submit" src="images/submit.png" />
            </section>



    </form>

Here is an example of the section of code that I'm trying to get to show when the submitted form data matches a certain value along with my script in relation:

    <div class="word" id="1" data-value="1">

<img src="images/Bat.png" class="letter_text">

<form id="f1" action="results.php" method="post">
    <input type="hidden" name="child_name" value="<?php $child_name; ?>" />
    <input type="hidden" name="running_score" value="<?php $running_score; ?>" />
    <input type="hidden" name="last_pic_displayed" value="1" />
    <input type="hidden" name="correct_spelling" value="BAT" />
    <input type="hidden" name="last_pic_flag" value="NO" />
    <input type="hidden" name="level" value="3_letter" />
    <input type="text" name="submitted_spelling" />
    <input type="image" class="submit" src="images/submit.png" />
</form>

</div>

<script>
    var lpic = "<?php echo $last_pic_displayed; ?>";
    lpic = parseInt(lpic);
    if (lpic == 0) {
        $("#1").show();
    }
    alert(lpic);
</script>

I'm pretty new at all this, so I definitely appreciate your help.

EDIT: Updated with syntax corrections; problem currently persists.

EDIT2: Updated with form example.

EDIT3: Updated with parseInt() and example of the div I'm trying to show.

Upvotes: 0

Views: 98

Answers (3)

geetanjali vaishnav
geetanjali vaishnav

Reputation: 31

use it:-

    var lpic = "<?php echo $last_pic_displayed; ?>";
lpic = parseInt(lpic);
if (lpic == 0) {
    $("#1").show();
}

Upvotes: 1

Lakhan
Lakhan

Reputation: 13226

You need to parseInt() the "lpic" variable Like that way :

var lpic = "<?php echo $last_pic_displayed; ?>";
lpic = parseInt(lpic);
if (lpic == 0) {
    $("#1").show();
}

Upvotes: 2

Manoj Dhiman
Manoj Dhiman

Reputation: 5166

use this 

<script>
    var lpic = "<?php echo $last_pic_displayed; ?>";  // this must in ""   
    if (lpic == '0') {                // must be == instead =
        $("#1").show();
    }
    </script>

Upvotes: 0

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