Make variable variable accessible outside of loop

Question

I am trying to make a looped variable variable accessible outside of the loop. Everything works, but I am having trouble with this piece of the code:

$dayVar."_count" = $dayVarCount;

The full code is below:

<?php

$day_1="sep_28";
$day_2="sep_29";
$day_3="sep_30";

$query = mysql_query("SELECT * FROM table WHERE id = '$id'");

while ($row = mysql_fetch_assoc($query))
{

for ($i = 1; $i <= 3; $i++) 
    {
        $dayVar = "day_".$i;
        $dayVarCount = $row[$$dayVar];
        echo $$dayVar . ': ' . $dayVarCount . '<br>';

        $dayVar."_count". = $dayVarCount;
    }      
}

echo "$day_3_count";

?>

Please describe what is your trouble. IE: what you actually have, and what your expect to have. By the way, the expression $dayVar."_count". = $dayVarCount; cannot be a variable assignation. — Skrol29
– Skrol29, Commented Oct 1, 2012 at 7:48
Could you please provide us with the error message you are getting? — LuigiEdlCarno
– LuigiEdlCarno, Commented Oct 1, 2012 at 7:48
What is the significance of $day_1="sep_28";? is this important to what you're trying to do - I think that the answers might change if so. — HorusKol
– HorusKol, Commented Oct 1, 2012 at 8:10

10cool · Accepted Answer · 2012-10-01 07:50:53Z

2

i believe that your variable assignment of of $dayVar."_count" is incorrect or illegal. i would try ${$dayVar."_count"} instead.

answered Oct 1, 2012 at 7:50

10cool

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1 Comment

LuigiEdlCarno Over a year ago

Wow, I didn't know you could do that in php.

LuigiEdlCarno · Accepted Answer · 2012-10-01 07:53:35Z

0

As Skrol29 rightly says, that line can not work:

An assignment (assigning a value to a veriable) always has the form

<variablename> <assignmentoperator> <value>

in php that means:

$day_1 = "123"

as you have already used.

Variable names are not usually not generic. If you need such functionality, use arrays. PHP arrays

answered Oct 1, 2012 at 7:53

LuigiEdlCarno

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Comments

BugFinder · Accepted Answer · 2012-10-01 07:54:37Z

0

You're almost in the rightplace.. but.

 for ($i = 1; $i <= 3; $i++)  
    { 
        $dayVar = "day_".$i; 
        $dayVarCount = $dayVar."_count";
    $$dayVarCount =  $row[$$dayVar];   
    }       


echo "$day_3_count";

answered Oct 1, 2012 at 7:54

BugFinder

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Comments

Sherin Jose · Accepted Answer · 2012-10-01 07:59:55Z

Try this,

 <?php
 // Connect to server and select database.
$host   =   "host here";
$username="username here";
$password="pass here";
$db_name="db here";
    mysql_connect("$host", "$username", "$password")or die("cannot connect");
   if(mysql_select_db("$db_name")){echo "Success";}else{echo "cannot select DB";}



   $day_1="sep_28";
   $day_2="sep_29";
   $day_3="sep_30";
   $id=1;
   $query = mysql_query("SELECT * FROM `tab1` WHERE id = '$id'");

   while ($row = mysql_fetch_assoc($query))
   {

   for ($i = 1; $i <= 3; $i++) 
   {
    $dayVar = "day_".$i;
    $dayVarCount = $row[$$dayVar];
    echo $$dayVar . ': ' . $dayVarCount . '<br>';

    ${$dayVar."_count"} = $dayVarCount;
  }      
  }

  echo "$day_3_count";

 ?>


  <?php
   // close connection
  mysql_close();
  ?>

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