passing parameter from php(mysql_query result) to javascript function triggered by onchange event

Question

I have an issue that I can't solve (I don't know how cause I'm a newbie in php and js). I have a text field and I want on blur trigger a function in js. My problem is that I want to pass a parameter in tnis function, which will get from a query. My test code is :

<script type="text/javascript">
  function calcNums(nums){
    var inText = "00.00";
    if (document.expences.mytext.value != inText){
    var mp = parseInt(nums.substr(0,2));
    var lp = parseInt(nums.substr(3,2));
    document.expences.mytext2.value = lpdeh*parseInt(document.expences.mytext.value)/100; 
    document.expences.mytext.value = mpdeh*parseInt(document.expences.mytext.value)/100; 
    } 
  }
</script>

<?php

    $selNums = "SELECT DEH_NUMS FROM BUILDING WHERE idBUILDING = '3'";
    $nums = mysql_query($selNums) or die("Could not execute query.");
    <input type="text" name="mytext" id="mytext"  value="00.00" onblur="calcNums($nums)"/>
    <input type="text" name="mytext2" id="mytext2"  value="00.00"/>
?>

My point is : I get $nums that is of type '50-50' (represents percentages). Get the initial value of "mytext" and calculate new values for "mytext" and "mytext2". example : $nums='60-40' initialMytext=100 --> new mytext=60 mytext2=40

If I put an inner var in calcNums() calculations are correct, but with the parameter is not triggered.

Can someone help? Thanks in advance!

I want to know how I can pass the parameter to event onblur="calcNums($nums) — Olga Anastasiadou
– Olga Anastasiadou, Commented Jan 23, 2011 at 16:30

shankhan · Accepted Answer · 2011-01-23 16:56:46Z

1

write this and it will fix

<script type="text/javascript">
  function calcNums(nums){
    var inText = "00.00";
    if (document.expences.mytext.value != inText){
    var mp = parseInt(nums.substr(0,2));
    var lp = parseInt(nums.substr(3,2));
    document.expences.mytext2.value = lpdeh*parseInt(document.expences.mytext.value)/100; 
    document.expences.mytext.value = mpdeh*parseInt(document.expences.mytext.value)/100; 
    } 
  }
</script>

<?php
    $selNums = "SELECT DEH_NUMS FROM BUILDING WHERE idBUILDING = '3'";
    $results = mysql_query($selNums);
    if (!$results) die("Could not execute query.");

    $row = mysql_fetch_assoc($results);
    $nums = $row['DEH_NUMS'];
?>
<input type="text" name="mytext" id="mytext"  value="00.00" onblur="calcNums('<?php echo $nums;?>')"/>
<input type="text" name="mytext2" id="mytext2"  value="00.00"/>

edited Jan 23, 2011 at 16:56

answered Jan 23, 2011 at 16:38

shankhan

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5 Comments

Olga Anastasiadou Over a year ago

I did this and instead of "60-40" that brings back my query, it gets "20" now.

shankhan Over a year ago

@Olga: I think you have few errors in your code is well I am editing my reply check it out I think it will help

Olga Anastasiadou Over a year ago

I wrote your code but results are the same. My query brings $nums='60-40' but when I put alert(nums) in calcNums, I get '20'! How is this possible????

shankhan Over a year ago

Are you using '<?php echo $nums;?>'... SINGLE QUOTATION... it should make it string and you should get nums as string in data. Can you check your page source what you are getting in browser for onblur="calcNums('???')"

Olga Anastasiadou Over a year ago

Sorry and thanks for your time and help. I didn't notice quotes! how stupid I am???

Satya · Accepted Answer · 2011-01-23 16:31:48Z

0

You're mixing HTML in PHP. Do something like this:

<?php
$selNums = "SELECT DEH_NUMS FROM BUILDING WHERE idBUILDING = '3'";
$nums = mysql_query($selNums) or die("Could not execute query.");
?>
<input type="text" name="mytext" id="mytext"  value="00.00" onblur="calcNums(<?php $nums ?>)"/>
<input type="text" name="mytext2" id="mytext2"  value="00.00"/>

answered Jan 23, 2011 at 16:31

Satya

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4 Comments

Olga Anastasiadou Over a year ago

Inputs are off php block. I tried "calcNums(<?php $nums ?>)" (this I didn't know before) and it gives me an undefined field when I put alert(nums) in my function. Can you tell me why?

Olga Anastasiadou Over a year ago

gives me "Resource id #9". Don't know what does this mean. In mysql the query returns the correct result.

Olga Anastasiadou Over a year ago

$selNums = "SELECT DEH_NUMS FROM BUILDING WHERE idBUILDING = '3'"; $res = mysql_query($selNums) or die("Could not execute query."); $row = mysql_fetch_array($res); $nums = $row['DEH_NUMS']; I found out what was wrong. This sould be my code. Thanks

Harish Over a year ago

@Olga Anastasiadou gives me "Resource id #9" actually what you get after mysql_query is result set you have to parse it using any of provide methods like while($row=mysql_fetch_array($res)) { $return[] = $row; }

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