CODEWITHSUNDEEP
php
MSD
MSD

Reputation: 349

Php variable in a variable

I have a piece of code that looks like this:

$result = mysql_query($queryc) or die(mysql_error());   

if(mysql_num_rows($result) > 0)                             
{   
  while($row = mysql_fetch_array($result, MYSQL_ASSOC))
  {
    echo $row['$field']; 
  }
}

Say that code was in a function and I wanted to pass $field to the $row[''] how would I accomplish that?

In other words I'm attempting to use $row['$field']; and $field is defined elsewhere

Upvotes: 2

Views: 164

Answers (4)

Ignacio Vazquez-Abrams
Ignacio Vazquez-Abrams

Reputation: 798646

Single quotes inhibit variable substitution.

echo $row["$field"];

or just

echo $row[$field];

The latter is highly recommended as it does not require PHP to parse $row["$field"] into $row[$field]. Saves you some microtime in each iteration.

Upvotes: 3

Gumbo
Gumbo

Reputation: 655239

Variables are not expanded in single quotes; they are only expanded in double quotes or in the heredoc syntax:

When a string is specified in double quotes or with heredoc, variables are parsed within it.

So either use double quotes or, even better, just omit them:

$row[$field]

Upvotes: 3

Neigyl R. Noval
Neigyl R. Noval

Reputation: 6038

suppose you have this function definition: 

function foo($field) {
    // from that code you have given
    ....
    echo $row[$field];  // no need of the quotation marks around $field
    ....
}

Upvotes: 4

Ben
Ben

Reputation: 62384

You'd not put any quotes around $field... like this:

echo $row[$field];

Upvotes: 4

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