Append PHP code into jQuery. [duplicate]

Question

I have a jquery function include html select and input, it shows form with selectable value name is olcu

<script type="text/javascript">
$(document).ready(function() {
    $("#add").click(function() {
        var intId = $("#buildyourform div").length + 1;
        var fieldWrapper = $("<div class=\"fieldwrapper\" id=\"field" + intId + "\"/>");
        var fMiktar = $("<input type=\"text\" class=\"miktar\" name=\"miktar[]\" size=\"6px\" />");
        var fOlcu = $("<select class=\"olcu\" name=\"olcu[]\"><option value=\"Adet\">Adet</option><option value=\"Metre\">Metre</option><option value=\"Gram\">Gram</option></select>");
        var fMalzeme = $("<input type=\"text\" name=\"malzeme[]\" class=\"malzeme\" id=\"aramayap\" />");
        var removeButton = $("<input type=\"button\" class=\"remove\" value=\"-\" />");
        removeButton.click(function() {
            $(this).parent().remove();
        });
        fieldWrapper.append(fMiktar);
        fieldWrapper.append(fOlcu);
        fieldWrapper.append(fMalzeme);
        fieldWrapper.append(removeButton);
        $("#buildyourform").append(fieldWrapper);
    });

});
</script>

I want to take values of olcu from database

And here is the database code

<script type="text/javascript">
    function showSelected(val){
        document.getElementById ('selectedResult').innerHTML = "" + val;
    }

</script>

<select name="olcu[]" onChange='showSelected(this.value)' class="olcu" id="olcu">Olcu Sec/option>

<?php

while($olculer=mysql_fetch_array($hangiolcu)){
echo "<option value='$olculer[olcu]'>$olculer[olcu]</option>";
}
?>
</select>

How can I append or replace that sql code into this jquery line

var fOlcu = $("<select class=\"olcu\" name=\"olcu[]\"><option value=\"Adet\">Adet</option><option value=\"Metre\">Metre</option><option value=\"Gram\">Gram</option></select>");

Why don't you use better jQuery selector for elements? by giving class name or ID ? — Ravi Dhoriya ツ
– Ravi Dhoriya ツ, Commented May 19, 2014 at 7:06
this looks strange (mix od javascript and php code), why dont you get data with ajax? — Bojan Kovacevic
– Bojan Kovacevic, Commented May 19, 2014 at 7:08
Your Database Code is not database code. It simply prints options from your database query. What you actually want to achieve is very unclear. What is that jQuery line supposed to do? I think you might have a XY Problem. — Andresch Serj
– Andresch Serj, Commented May 19, 2014 at 7:14

developer10 · Accepted Answer · 2014-05-19 07:15:55Z

2

You can create Javascript/jQuery variable with PHP code like this:

var foo = '<?php //echo or return php value here; ?>';

answered May 19, 2014 at 7:15

developer10

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1 Comment

Seyhan Over a year ago

Can you please combine code?

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