Insert PHP code in JQuery

Question

Good day every one. I want to insert my php code in my jquery. I want to show my data in database using PHP and I want to put it in a option box. I used var in jquery plus the code of my php but isn't working. Please help me.
Shows the data:

 <?php
 $res2 = mysql_query("SELECT * FROM expense_maintenance ORDER BY name Asc");
 while ($result2 = mysql_fetch_assoc($res)){
 $name = $result2["name"];
 ?>

jquery code(dynamic adding text box)

 var nitem =0; 
 var ntotal = 0;
 var option = <?php echo" <option value='$name'>$name</option>";} ?>;
 function totalItemExpence(){
 ntotal = 0;
 $('.expense_cost').each(function(){ 
 if($(this).val() != ""){
 ntotal += parseFloat($(this).val()); 
 }
 }); 
 //$('#total').val(ntotal); 
 }
 $(document).on('change keyup paste', '.expense_cost', function() { 
 totalItemExpence();
 mytotal();
 }); 
 $('.btn').click(function() { 
 nitem++; 
 $('#wrapper').append('<div id="div' + nitem + '" class="inputwrap">' +
 '<select class="expense_name" id="' + nitem '">"'+ option +'"</select>' +
 '<input class="expense_desc" placeholder="Expense Description" id="' + nitem + '" required/>' +
 '<input class="expense_cost" onkeypress="return isNumber(event)" placeholder="Expense Cost" id="' + nitem + '" required/> ' +
 '<br><br></div>'); 
 }); 
 $('.btn2').click(function() { 
 ntotal = $('#total').val(); 
 $("#div" + nitem + " .expense_cost").each(function(){ 
 if($(this).val() != ""){
 ntotal -= parseFloat($(this).val()); 
 }
 }); 
 $("#div" + nitem ).remove();
 nitem--; 
 $('#total').val(ntotal); }); 

• u_mulder: Check errors in developers console. Error obviously in var option definition.
• user3368390: I tried it for a while. I don't know how to use jquery. I'm a beginner. Sorry
• Patrick Geyer: You shouldn't be using mysql_ functions anymore as they've been deprecated. Use PDO or mysqli instead. They're more secure and it's very easy to convert your code
• u_mulder: I tried it for a while Tried what?
• user3368390: I am using this code var option = "<?php echo <option>$name</option>"; ?>"; but the loop is not working it not shows the full data of my field. it is showing the first data of my database. Can you help me.

Answer 1

Try this for while loop :

<?php
 $res2 = mysql_query("SELECT * FROM expense_maintenance ORDER BY name Asc");
 $options = '';
 while ($result2 = mysql_fetch_assoc($res)){
 $options .= "<option value='{$result2["name"]}'>{$result2["name"]}</option>";
 }
?>

And JS part :

...
var option = "<?php echo $options; ?>";
...

• user3368390: It think sir it has a problem in result2. And I don't know what should I insert.
• Nikola Loncar: In your example the loop has no effect. That's why you get only one result. First collect all the options and than put them to html via jquery.

Answer 2

This line:

var option = "<?php echo "<option value='$name'>$name</option>"; ?>";

must be in your .php file, because php code won't be executed in a .js file (that's why it's called .php)... You can wrap that line in a <style> tag in your .php file

Also, don't forget to add these things --> " And to remove this one --> }

• user3368390: I am using this code var option = "<?php echo <option>$name</option>"; ?>"; but the loop is not working it not shows the full data of my field. it is showing the first data of my database. Can you help me.
• sampie777: You better create a js array var option = new Array(); and declare this array before the php-code (the loop).Declare a $i=0; in php before the loop. In the loop you put: echo "option[$i]='<option value=\'$name\'>$name</option>';"; $i++; Because, the reason why you only get one row of your database, is because you rewrite var option again and again.