php textfield arrayname to js. need help

kav

while($rowdef=mysql_fetch_array($resultdef))
{
$app_criteriaid=$rowdef['app_criteriaid'];
$app_criteria=$rowdef['app_criteria'];
echo "<tr bgcolor=\"#EBF5FF\"><td><div align=\"center\">$no</div></td>";
echo "<td>".$rowdef['app_criteria']."</td>";
echo "<td><input type=\"text\" name=\"arraypoint[]\" value=\"\"></td>";
echo "<td><input type=\"text\" name=\"arrayremark[]\" value=\"\"></td>";
$no++;
}

i name the name as arraypoint[] because when i submit the form, i will only need to retrieve the array to have all the values. Since when querying the database the result always follows an order, i pass the value in it assuming its 1 to 1. (exact match)

The problem now is i want to check whether the input is a numeric before submitting using javascript.
i tried using this but it seems nothing happen

Weedpacket

http://www.php.net/manual/en/faq.html.php#faq.html.arrays
Note that if you are using JavaScript the [] on the element name might cause you problems when you try to refer to the element by name. Use it's numerical form element ID instead, or enclose the variable name in single quotes and use that as the index to the elements array, for example:
variable = documents.forms[0].elements['var[]'];

kav

since mine is this:
document.formname.arraypoint.value

what must be put inside this ['var[]']??
documents.forms[0].elements['var[]'];

simplier...var[]=what?

Weedpacket

Basic Javascript this. Form objects have an array called "elements" containing all the form's elements. This array has both numeric and associative indices. The numeric indices follow the same order as elements appear in the form. The associative indices are the names of the form elements. Since in your case the name of the form element is "arraypoint[]" (because that's what you've called it in the field's name= attribute), the whole DOM reference is document.formname.elements['arraypoint[]'].
In other words,

...enclose the variable name in single quotes and use that as the index to the elements array...

But personally I use document.getElementById().

kav

thanks alot, Weedpacket.
However, I would like to know about the one u said u personally use which is document.getElementById().
How do u name it? <input name=? or issit id=? ......

after that how to refer to it in javascript? can give example.

kav

opss..here is it
function validateForm()
{

var thenumber=document.appraisal_form.elements['arraypoint[0]'];

document.write(thenumber[0]);

}

is this the way? i tried to display it but nothing appear. if i put document.write(thenumber); without the bracket, it display [object]

document.write(document.appraisal_form.elements['arraypoint[0]']); it displays undefined

Anyway, can explain this sentence and provide some examples. its from previous post.=>
Use it's numerical form element ID instead, or enclose the variable name in single quotes and use that as the index to the elements array😕

Weedpacket

Something like this test:

<html>
<head><title></title></head>
<body>
<form>
<input type="text" name="field1[]" value="foo">
<input type="text" name="field1[]" value="bar">
<input type="text" name="field1[]" value="baz">
<input type="text" name="field1[]" value="wibble">
<input type="button" value="Test" onClick="alert(document.forms[0].elements['field1[]'][2].value)">
</form></body></html>

tomandhannah

This last post helped me a lot. I have a form with dynamcially generated drop-down menus numbering 1 through no. of records from MySQL. But I was using a JavaScript onChange to automatically update all the drop-down menus when one is changed. For example: It prints a list of 9 drop-down menus all valued at 1-9 respectively. Then, if you change the first one from "1" to "2", it automatically updates the second one from "2" to "1"...

Well, I wanted to submit these new values as an array using name="menu[]" but that kept my JavaScript from working. So, instead I replaced my onChange from this:
onChange="reOrder(menu[], <? echo $t++; ?>);"

To this:
onChange="reOrder(document.forms[0].elements['menu[]'], <? echo $t++; ?>);"

And now it works perfectly!