php for loop creating undefined variable for string values

Question

I am looping though my data and strong the result in a variable. It is currently being stored as strings.

for ($i = 0; $i < count($AnotherArray); $i++){

  $myArray  .= $dataArray[$i].",";
 }

This correctly returns the content with commas separating the values. But i get a notice

Notice: undefind variable: $myArray in ...

The above is the first time i create and call $myArray

$myArray = .= , just .= would be enough and you should declare an string variable before the loop to avoid the warning $myArray = ""; — Frankich, Aug 25 '17 at 15:18
Should be `$myArray .= $dataArray[$i].",";` you have an extra equals sign that is a syntax error — GrumpyCrouton, Aug 25 '17 at 15:18

crazyloonybin · Answer 1 · 2017-08-25T15:41:05.947

You need to declare the $myArray variable first. So your code becomes:

$myArray = '';
for ($i = 0; $i < count($AnotherArray); $i++){
    $myArray .= $dataArray[$i].",";
}

As a side note, I'd also recommend looking at the naming of your variables, $myArray is actually a string, not an array. Also, $dataArray and $AnotherArray don't describe the data the variables are storing. When coding, it is useful to give variables meaningful names, so that yourself and others who may look at the code will find it easier to follow.

score 2 · Accepted Answer · answered Aug 25 '17 at 15:21

2

You need to have the variable defined before you can append to it

$myArray="";
for ($i = 0; $i < count($AnotherArray); $i++){

  $myArray  .= $dataArray[$i].",";
 }

answered Aug 25 '17 at 15:21

Carlos Alves Jorge

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php for loop creating undefined variable for string values

2 Answers2