How to create a php variable with text and value of another variable

Question

Want to create a php variable using text before and after the value of another variable.

variable variables. But have only seen examples of assignment with no text.

$vsBOA_W[]=$rows['vsBOA_W'];

// BOA = team 3-char abbreviation. Looking for something similar to above but insert 3-char abbreviations based on a input file.

$numOfTeams = 3;    // Determined from external source

$teamAbbr = array("BOA","CAA","CHN");   // For simplicity for this example. This array would normally be created from an external source. 

for($i=0; $i<$numOfTeams; $i++) {   // I know I can use size of array instead of nunOfTeams. That's not the issue. 

    echo $teamAbbr[$i]."<br>";      // for testing

    $$("vs".{'$teamAbbr[$i]'}."_W[]"} = $rows['$$("vs".{'$teamAbbr[$i]'}."_W"}']; // a total guess
}

I expect the end result to look like:

$vsBOA_W[]=$rows['vsBOA_W'];

for BOA

Update #2: I tried the following (breaking down each step) and get the same error on $$TeamWins assignment.

for($i=0; $i<$numOfTeams; $i++) {    
    echo $teamAbbr[$i]."<br>";      
    $TeamWins = 'vs' . $teamAbbr[$i] . '_W';
    echo "TeamWins=$TeamWins<br>";
    $TeamWinsHold = $rows[$TeamWins];
    echo "TeamWinsHold=$TeamWinsHold<br>";
    $$TeamWins[] = $TeamWinsHold;
}

Update #3:

for($i=0; $i<$numOfTeams; $i++) {    
    echo $teamAbbr[$i]."<br>";      
    $TeamWins = 'vs' . $teamAbbr[$i] . '_W';
    echo "TeamWins=$TeamWins<br>";
    $TeamWinsHold = $rows[$TeamWins];
    echo "TeamWinsHold=$TeamWinsHold<br>";
    ${$TeamWins}[] = $TeamWinsHold;
}

foreach(${$TeamWins} as $value) {
    echo "value=$value<br>";   // only displays last element or value assigned from above loop.
}

Update #4 (final):

$teamW = array();
$teamL = array();
for($i=0; $i<$numOfTeams; $i++) {    
    //echo $teamAbbr[$i]."<br>";        
    $teamWName = 'vs' . $teamAbbr[$i] . '_W';
    $teamLName = 'vs' . $teamAbbr[$i] . '_L';
    //echo "teamWName=$teamWName<br>";
    //echo "teamLName=$teamLName<br>";
    $teamW[$teamWName] = $rows[$teamWName];
    $teamL[$teamLName] = $rows[$teamLName];
}

For clarification, I wish to make this code generic for multiple tables. One table has a set of teams (ex: BOA,CAA,CHN), another table has another set of teams (ex: ABC,DEF,GHI). The external file tells the php code which teams to process (could be all or just a few). — user2140857
– user2140857, Commented Jul 27, 2019 at 4:14

Progrock · Accepted Answer · 2019-07-27 18:02:21Z

1

I don't quite understand the interplay with the rows in your example. But going by your guess assignment, you can always simplify, by forming the variable name upfront:

<?php

$rows  = ['xFOOy'=>[], 'xBARy'=>[], 'xBAZy'=>[]];
$items = ['FOO', 'BAR', 'BAZ'];

foreach($items as $abbr)
{
    $name = 'x' . $abbr . 'y';
    ${$name}[] = $rows[$name];
}

But, I'd say you'd be better off with a keyed array than variable variables, as it makes for easier inspection, and there is less chance of namespace clashes.

edited Jul 27, 2019 at 18:02

answered Jul 27, 2019 at 4:21

Progrock

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10 Comments

user2140857 Over a year ago

The reasons for the rows reference, $rows['vsBOA_W'], is that I'm looping through all the years / records for a team and displaying in a form as an array, $vsBOA_W[], which can be updated.

user2140857 Over a year ago

The $$name[] = $rows[$name]; returns an error:Cannot use [] for reading

user2140857 Over a year ago

See my Update #2.

Progrock Over a year ago

@user2140857, you must be using an end of life PHP version, edited to address that error.

user2140857 Over a year ago

Correct. Using PHP 5.6. It's not production code. Never will be. Never had a requirement to go 7.0. Maybe I do now?

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