Question

I seem to have looked everywhere for this.

Here is the code i use to set a variable to a piece of data from a MySQL database

$PartNo=mysql_result($result,$i,"PartNo");

Where

 $result = mysql_query("SELECT * FROM PriceList"); 

$i = 0, is added to by 1 every time my while loop restarts
PartNo is a field name in my MySQL table and also the name of the variable I want to set the data in the database to
PriceList is my database name

I want to loop through all the field names (the array has them) and set variables with the same names to that data. Like this:

$PartNo=mysql_result($result,$i,"PartNo");
$Group=mysql_result($result,$i,"Group");
$OnHand=mysql_result($result,$i,"OnHand");
$QOO=mysql_result($result,$i,"QOO");
$Description=mysql_result($result,$i,"Desc");
$Cost=mysql_result($result,$i,"Cost");

But with a foreach loop so it isn’t as much code.
I was thinking something like this, but it won’t execute no matter which way I go about it (parse_str, eval, exec, etc.)

$nDatabaseVars=array("PartNo","Group","OnHand","QOO","Desc","Cost");

foreach ($nDatabaseVars as $X) {
    $$X=mysql_result($result,$i,'$X');
}

I need “$$X” to evaluate out so on the first iteration, it changes to $PartNo= and then sets $PartNo to whatever data is in the database on the first line. Which is what this part is: mysql_result($result,$i,"PartNo")

If I echo it out instead:

foreach ($nDatabaseVars as $X) {
    echo "$$X=mysql_result($result,$i,'$X')";
}

I can get it to say exactly what I need executed ($PartNo=mysql_result($result,$i,”PartNo”);) but not actually get the variable set.

Answer 1

You are passing a string containing "$X" to mysql_result, not the name of your column. Remove the single quotes:

$$X=mysql_result($result, $i, $X);