Use for loop to incrementally select values from MySQL query

Question

I am trying to use a for loop to select an incremental value with a MySQL query. I have included sample code 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 = $dayVar."_count"; // Don't really need this anymore, so removed.
        $dayVarCount = $row[$$dayVar];
        echo "$$dayVar.': '.$dayVarCount<p>"; // Edited.
    }      
}
?>

I think I am getting close, but when I run the code my page is showing this:

$day_1.': '.0

$day_2.': '.2

$day_3.': '.5

Any additional recommendations? Thanks for the great help!

Can you explain a little more? I don't understand what you're trying to accomplish here. Running queries in a for loop can be pretty inefficient. If there's anyway to accomplish it with a single sql query, you will be much better off. — Adam Plocher
– Adam Plocher, Commented Oct 1, 2012 at 6:36
What exactly do you try to display? Do you want to count the days of the selected month? — nja
– nja, Commented Oct 1, 2012 at 6:37

Stegrex · Accepted Answer · 2012-10-01 14:23:13Z

1

Try a variable variable:

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

This basically uses a string to reference a variable by its name.

Just think of it this way:

$variable = 'hello';

$string = 'variable';

echo $$string;
// Is the same thing as:
echo $variable;

// Because you can thing of $$string as ${$string} ---> $variable when {$string} is interpreted into 'variable'

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

edited Oct 1, 2012 at 14:23

answered Oct 1, 2012 at 6:40

Stegrex

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7 Comments

Brandon Over a year ago

Thanks stegrex, I looked at some of the documentation for this and thought it might be the right path, I will try this code out. THANK YOU!

Stegrex Over a year ago

@Brandon Sure thing, and good luck! You can let me know if it works or not and I can update my answer.

Brandon Over a year ago

Thanks for the explanation @Stegrex! I have updated the code above, I think I am getting close.

Stegrex Over a year ago

@Brandon I've made some edits to my answer. Basically, not having the variable variable be interpolated in the double quotes.

Brandon Over a year ago

Ive implemented the code and think I am getting closer here, please see above. Also, I added an additional double quote at the beginning of the "echo" line as there wasn't one. THANKS!

|

Mahdi · Accepted Answer · 2012-10-01 07:05:07Z

1

Replace this line:

echo "$$dayVar.': '.$dayVarCount<p>";

with this:

echo $$dayVar . ': ' . $dayVarCount . '<br>';

answered Oct 1, 2012 at 7:05

Mahdi

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2 Comments

Brandon Over a year ago

Yep, @Stegrex had given me all that code, but your adjustment made it all click. Thank you very much!

Mahdi Over a year ago

Glad to hear that, yes @Stegrex solution is fitting I guess. I upvote that! :D

Hemi · Accepted Answer · 2012-10-01 06:52:30Z

0

<?php

$day_1="January 1";
$day_2="January 2";
$day_3="January 3";

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

while ($row = mysql_fetch_assoc($query))
{
for ($i = 1; $i <= 3; $i++) 
    {
        $var = $."day_".$i;
        $day_$i_count=$row['$var'];
        echo "$day_$i: $day_$i_count<p>";
    }       
}
?>

You can try this code too.

answered Oct 1, 2012 at 6:52

Hemi

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Comments

FoolishSeth · Accepted Answer · 2012-10-01 07:06:58Z

0

That all seems overly complicated, but possibly I don't understand the question adequately. Would this work?

$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))
{
    foreach ($day as $day_str) 
    {
        echo $day_str . ':' . $row[$day_str] . '<p>';
    }      
}

answered Oct 1, 2012 at 7:06

FoolishSeth

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