PHP Variable Variable for loop

Question

hey guys is this even possible to be done in php

<?php   
$sectorcount = $row_SectorCount['COUNT(*)'];
//number of rows in database
for ($i = 1; $i <= $sectorcount; $i++)
{
${spendingname . $i}= array();
${spendingpercent . $i} = array();
${spendingid . $i} = array();

mysql_select_db($database_conn2, $conn2);
$query_Spending = "SELECT CONCAT(spending.SectorID, spending.ExpenditureID) AS 'SpendingID',
expenditure.ExpenditureName, spending.SpendingPercent, spending.SectorID
FROM spending   
INNER JOIN expenditure ON spending.ExpenditureID = expenditure.ExpenditureID
WHERE spending.SectorID = ".$i;
$Spending = mysql_query($query_Spending, $conn2) or die(mysql_error());
$totalRows_Spending = mysql_num_rows($Spending);
while($row_Spending = mysql_fetch_assoc($Spending))
{
${spendingname.$i}[] = $row_Spending['ExpenditureName'];
${spendingpercent.$i}[] = $row_Spending['SpendingPercent'];
${spendingid.$i}[]= $row_Spending['SpendingID'];
}
mysql_free_result($Spending);
}

i was planning to use this here in this context. having an array to draw out the values using a where clause and display them individually

Levi Morrison · Accepted Answer · 2012-01-04 02:27:15Z

3

This is what you are looking for:

for ($i = 0; $i < 14; $i++) {
    ${'name' . $i} = $i;
}

echo $name3;

But you should just use an array:

$name = array();
for ($i = 0; $i < 14; $i++) {
    $name[$i] = $i;
}

echo $name[3];

If you need an array in an array, go ahead and do that too. If you need more nested arrays than that, you should probably be using a class somehow instead of all that array nesting.

edited Jan 4, 2012 at 2:27

answered Jan 4, 2012 at 2:03

Levi Morrison

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6 Comments

Jervis Chionh Over a year ago

thank you!! i want to try it on an array within an array like $name.$i = array (); then using database to draw the data out

Levi Morrison Over a year ago

@JervisChionh You should upvote answers you find useful. You should also accept an answer to your question by clicking on the check mark. Welcome to StackOverflow.

Jervis Chionh Over a year ago

i have in sufficient rep ! i would love to do that

Jervis Chionh Over a year ago

heyy levi what if i want something like that $spendingname = array(); for($i=1; $i<=13; $i++) { $spendingname[$i] = one,two,three; } echo $spendingid[1]; ?>

Levi Morrison Over a year ago

@JervisChionh I'm confused by your code. You use both $spendingname and $spendingid.

|

Jon · Accepted Answer · 2012-01-04 02:03:00Z

2

First off, I second the suggestion to use an array instead. Variable variables are not meant to be used like this.

That said, it is possible by accessing ${$name.$i}, or perhaps doing something like

$varname = $name.$i;
$$varname = 1; // or read from $$varname

answered Jan 4, 2012 at 2:03

Jon

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Comments

Niet the Dark Absol · Accepted Answer · 2012-01-04 02:00:39Z

1

Use an array. That's what they're there for.

answered Jan 4, 2012 at 2:00

Niet the Dark Absol

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3 Comments

Jervis Chionh Over a year ago

what if i want to have an array within an array?

Niet the Dark Absol Over a year ago

Then just do so. Nothing wrong with arrays in arrays in arrays in arrays in arrays.

Tomas Reimers Over a year ago

in arrays in arrays. They are called nested arrays.

Tomas Reimers · Accepted Answer · 2012-01-04 02:04:28Z

1

As Kolink said, that is what an array is for!

However, What you are asking can be achieved:

Comments

score 1 · Accepted Answer · 2012-01-04 02:11:13Z

1

It will not give desired output 1 instead will give 13

What s the reason of doing like this. you can make array a access is later on use

 for($i=0; $i<=13; $i++)
 {
        $name[$i]=$i;
 }

      echo $name[1];                    *//it will echo 1

edited Jan 4, 2012 at 2:11

answered Jan 4, 2012 at 2:05

user319198

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