problem with enable/disable button using javascript

Question

Im am trying to add a check box that will enable/disable the edit button. Im retrieving a price list and displaying it inside a table. When i add the javascript into the php code it doesn't work. Below is my code

<table border="1">
  <tr>

    <td width="100">Fee % </td>
    <td width="100">Price</td>
    <td width="100">Total</td>
    <td width="102">&nbsp;</td>

  </tr>
  <tr>
    <?php
  $sql1="select * from pricelist";
  $result1=mysql_query($sql1) or die(mysql_error());

  while ($row=mysql_fetch_array($result1)) {

$id=$row['id'];
$price=$row['h_price'];
$a=0;

print "<form id='form1' name='$a+' method='post' action=''>";
print "<td><input name='fees' value ='$fees' type='text' size='4' /></td>";
print "<td><input name='price' value ='$price' type='text' size='15' /></td>";
echo "<td><input type='checkbox' onclick='this.$a+.disabled = !this.checked;'><td>";
print"<td><input type='submit' name='$a+' value='Submit' disabled='disabled' /></td>"; 
print "</tr>";
print "</form>";

    }
   ?>
</table>

Can someone please tell me what am i doing wrong?

Thanks

KoolKabin · Accepted Answer · 2010-06-07 09:30:57Z

1

take out the following code out of loop... it creates multiple form tag...

print "<form id='form1' name='$a+' method='post' action=''>";

Place "<tr></tr>" inside the loop...

final code looks like:

<form id='form1' name='fm1' method='post' action=''>  <tr>
    <?php
  $sql1="select * from pricelist";
  $result1=mysql_query($sql1) or die(mysql_error());

  while ($row=mysql_fetch_array($result1)) {

$id=$row['id'];
$price=$row['h_price'];
$a=0;
print "<tr>";
print "<td><input name='fees' value ='$fees' type='text' size='4' /></td>";
print "<td><input name='price' value ='$price' type='text' size='15' /></td>";
echo "<td><input type='checkbox' onclick='this.$a+.disabled = !this.checked;'><td>";
print"<td><input type='submit' name='$a+' value='Submit' disabled='disabled' /></td>"; 
print "</tr>";

    }
print "</form>";
   ?>

Next if u want to control the button only then

<input type='checkbox' onclick='document.$a.submit.disabled = !this.checked;' />

and make sure of the following things: 1.) form name should be $a i.e <form name='$a' ...>

2.) submit buttons name should be submit i.e <input type='submit' name='submit'...>

3.) increase the $a variable only at the end of loop i.e

$a++;
}

edited Jun 7, 2010 at 9:30

answered Jun 7, 2010 at 9:21

KoolKabin

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2 Comments

LiveEn Over a year ago

But im still having the javascript problem. Once i tick the checkbox the button doesn't get enabled..

KoolKabin Over a year ago

it should work fine if u have given attention to my points 1-3... here is a static eg: <form name="f1"> <input type='checkbox' onclick='document.f1.submit.disabled = !this.checked;' /> <input type="button" name="submit" value="kool" disabled="disabled" /> </form>

Adeel · Accepted Answer · 2010-06-07 09:04:55Z

1

i think $a+ will cause syntax error. you need to increment $a at the end of loop like $a++. Also see page source and see what is coming from server.

answered Jun 7, 2010 at 9:04

Adeel

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Comments

user58670 · Accepted Answer · 2010-06-07 09:15:46Z

1

I would recommend using following format for html + php output

<form id='form1' name='<?=$a++?>' method='post' action=''>
<tr>
<td><input name='fees' value ='<?=$fees?>' type='text' size='4' /></td>
<td><input name='price' value ='<?=$price?>' type='text' size='15' /></td>
<td><input type='checkbox' onclick='this.disabled = !this.checked;'><td>
<td><input type='submit' name='<?=$a++?>' value='Submit' disabled='disabled' /></td>
</tr>
</form>
<?

you also need to have < tr > to be in the while loop.

answered Jun 7, 2010 at 9:15

user58670

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1 Comment

Strae Over a year ago

short tags will be deprecatd in php6

Andre · Accepted Answer · 2010-06-07 09:16:34Z

0

In your html part where you set up the formular you have to use " " to escape the php code. example:

print "<form id='form1' name='".$a+."' method='post' action=''>";

But as Adeel said already, what is $a+ ?

And you have to edit all prints and echoes in the way, that php expressions are excaped from the HTML source.

answered Jun 7, 2010 at 9:16

Andre

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