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Bug #10412 field name in mysql_lsit_fields
Submitted: 2001-04-19 18:59 UTC Modified: 2001-04-19 19:11 UTC
From: dkokenge at netscape dot net Assigned:
Status: Closed Package: Scripting Engine problem
PHP Version: 4.0.4pl1 OS:
Private report: No CVE-ID: None
 [2001-04-19 18:59 UTC] dkokenge at netscape dot net
When using "mysql_list_fields" for a file you can extract
the field names of the schema.
For example: $fname. Then turn it into the variable for the
schema name as $$fname.
$$fname = 'something';  // **** works
$$fname[0] = 'something'; // **** don't work
---------------------------------------------------
Example
---------------------------------------------------
$schema = mysql_list_fields("wine","wpo_dtl");	
$nf 	= mysql_num_fields($schema);	// get num of fields
$xx 	= mysql_fetch_array($schema);	
while ($i < $nf) {
$fname = mysql_fieldname($schema,$i);

//***** this works *****
$$fname = 'something';
// ******* this don't
$x = 0;
$$fname[$x] = 'something';
}

What am I doing wrong???


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 [2001-04-19 19:11 UTC] torben@php.net
Please read the following page in the manual, especially
the last paragraph:

http://www.php.net/manual/en/language.variables.variable.php

The problem is that $$fname[] is ambiguous and you need to
tell PHP what it means. The page above explains in more 
detail.
 
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