siric Posted November 5, 2009 Share Posted November 5, 2009 Hi, I am pulling data from a database to populate a form for editing purposes. The form includes many radio buttons. Is there a more elegant way to accomplish this than the following - if ($var1 = '1') { print "<input type='radio' value='1' checked name='quality'>1<p>; } else { print "<input type='radio' value='1' name='quality'>1<p>; } if ($var1 = '2') { print "<input type='radio' value='2' checked name='quality'>2<p>; } else { print "<input type='radio' value='2' name='quality'>2<p>; } ...... Thanks Link to comment https://forums.phpfreaks.com/topic/180451-solved-populate-form-fields-from-database/ Share on other sites More sharing options...
mrMarcus Posted November 5, 2009 Share Posted November 5, 2009 <?php echo '<input type="radio" value="1" name="quality"'.(($var1 = '1') ? ' checked' : '').'" />'; ?> and so on for each $var; Link to comment https://forums.phpfreaks.com/topic/180451-solved-populate-form-fields-from-database/#findComment-951957 Share on other sites More sharing options...
siric Posted November 5, 2009 Author Share Posted November 5, 2009 <?php echo '<input type="radio" value="1" name="quality"'.(($var1 = '1') ? ' checked' : '').'" />'; ?> and so on for each $var; Thanks. Will try this. Link to comment https://forums.phpfreaks.com/topic/180451-solved-populate-form-fields-from-database/#findComment-951969 Share on other sites More sharing options...
siric Posted November 6, 2009 Author Share Posted November 6, 2009 <?php echo '<input type="radio" value="1" name="quality"'.(($var1 = '1') ? ' checked' : '').'" />'; ?> and so on for each $var; Thanks. Will try this. Have tried it and it doe snot work. Button is always unchecked. Link to comment https://forums.phpfreaks.com/topic/180451-solved-populate-form-fields-from-database/#findComment-952649 Share on other sites More sharing options...
mrMarcus Posted November 6, 2009 Share Posted November 6, 2009 does $var1 or $var2 have a value of either 1 or 2? have you echo'd them out to make sure? post all relevant code (query, form). Link to comment https://forums.phpfreaks.com/topic/180451-solved-populate-form-fields-from-database/#findComment-952654 Share on other sites More sharing options...
mrMarcus Posted November 6, 2009 Share Posted November 6, 2009 my bad, try this: <?php echo '<input type="radio" value="1" name="quality"'.(($var1 == '1') ? ' checked' : '').'" />'; ?> i had given you the wrong comparable (= instead of ==) .. try this code now. EDIT: for the record, i rarely test code that i post. Link to comment https://forums.phpfreaks.com/topic/180451-solved-populate-form-fields-from-database/#findComment-952655 Share on other sites More sharing options...
siric Posted November 6, 2009 Author Share Posted November 6, 2009 my bad, try this: <?php echo '<input type="radio" value="1" name="quality"'.(($var1 == '1') ? ' checked' : '').'" />'; ?> i had given you the wrong comparable (= instead of ==) .. try this code now. EDIT: for the record, i rarely test code that i post. That was the problem. I started from an if statement from scratch and realised just before I saw this post. Much Thanks. Link to comment https://forums.phpfreaks.com/topic/180451-solved-populate-form-fields-from-database/#findComment-952721 Share on other sites More sharing options...
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