PHP illegal string offset error

Question

I have this code I don't know why I get error from it.

if( ! in_array($repeatType, ['monthly', 'weekly', 'daily'])){
    // do somehting
}
$monthly = ['two_years' => 26, 'offset_const' => 4, 'add_unite' => 'weeks'];
$weekly = ['two_years' => 52*2, 'offset_const' => 1, 'add_unite' => 'weeks'];
$daily = array('two_years' => 365*2, 'offset_const' => 1, 'add_unite' => 'days');

for ($i=0; $i < $$repeatType['two_years']; $i++) { #<--- here I get the error

    // ..... // rest of the code

This is so weird cause I have checked var_dump($$repeatType) output, it seems fine:

array(3){["two_years"]=>int(730)["offset_const"]=>int(1)["add_unite"]=>string(4)"days"}

@AbhikChakraborty it's fine to have double $$, since it makes variable $monthly, $weekly or $daily to be dynamicaly selected — Justinas
– Justinas, Commented Jul 14, 2014 at 7:28
@GuyT nope, $foo["two_years"] is a number, no count() missing. — Maerlyn
– Maerlyn, Commented Jul 14, 2014 at 7:30
@Maerlyn it caused syntax error syntax error, unexpected '[', expecting ';'. — Walid Ammar
– Walid Ammar, Commented Jul 14, 2014 at 7:31

Maerlyn · Accepted Answer · 2014-07-14 07:34:09Z

6

This is a syntax limitation. PHP is trying to bind the array-index operator to $repeatType (which is a string) and an associative key is invalid in a string, thus causing your problem.

You need to explicitly specify where your variable starts and begins like this:

for ($i=0; $i < ${$repeatType}['two_years']; $i++) {}

A workaround is to assign it to a temporary variable like this:

$selectedRepeatType = $$repeatType;
for ($i=0; $i < $selectedRepeatType['two_years']; $i++) {}

answered Jul 14, 2014 at 7:34

Maerlyn

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6 Comments

Walid Ammar Over a year ago

Thanks ... never heard about that before.

peter_the_oak Over a year ago

Note that there is not really a need for a workaround. The curly brackets as you mentioned work just fine.

Walid Ammar Over a year ago

@peter_the_oak I have tried curly brackets but I got a syntax error after using them.

peter_the_oak Over a year ago

I tried them too :-) Like this: 1. $name = "two_years"; 2. $two_years = array('A' => 10, 'B' => 20); 3. print_r(${$name}['A']); On my side, these lines work like a charm.

Walid Ammar Over a year ago

@peter_the_oak oh.. sorry you're right !!, I have tried normal brackets before which is mentioned in the comments above but curly brackets works!!

|

bixiaopeng · Accepted Answer · 2014-07-14 07:44:32Z

0

I can understand what you want ,but this expression $$repeatType['two_years'] is ambiguous , at least for me it takes me some time to run the code to find out what the output,so first of all I should suggest you don't use such a expression.

OK,come back to the question,you expected $$repeatType['two_years'] to return one of the number 26,52*2,and 365*2.But let us run the code.

$name='a';
$a=array('n'=>1);
$a=array('n'=>2);
$a=array('n'=>3);

var_dump($$name['n']);  //array(1) { ["n"]=> int(3) }
var_dump($$name);       //array(1) { ["n"]=> int(3) }
var_dump(${$name}['n']);//int(3)

OK ,we know what happend, and you can find why here.

answered Jul 14, 2014 at 7:44

bixiaopeng

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Comments

DC- · Accepted Answer · 2014-07-14 08:24:10Z

0

From your code, i can understand that you haven't set a value for $repeatType['two_years']. First assign a value

$repeatType['two_years'] = "monthly";

Now "$monthly" is an array so you call:

for ($i=0; $i < ${$repeatType['two_years']}['two_years']; $i ++){
    //Loop 26 times.
}

$repeatType['two_years'] need to be defined or it will give a PHP error.

answered Jul 14, 2014 at 8:24

DC-

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1 Comment

Walid Ammar Over a year ago

if you check my code again, you can realize that the first line of it checks for $reapeatType value, it's already assigned, the problem was I didn't add the curly brackets, thanks for your suggestion.

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