Cannot use string offset as an array in php

问题:

I'm trying to simulate this error with a sample php code but haven't been successful. Any help would be great.

"Cannot use string offset as an array"

回答1:

For PHP4

...this reproduced the error:

$foo    = 'bar'; $foo[0] = 'bar'; 

For PHP5

...this reproduced the error:

$foo = 'bar';  if (is_array($foo['bar']))     echo 'bar-array'; if (is_array($foo['bar']['foo']))     echo 'bar-foo-array'; if (is_array($foo['bar']['foo']['bar']))     echo 'bar-foo-bar-array'; 

(From bugs.php.net actually)

Edit,

so why doesn't the error appear in the first if condition even though it is a string.

Because PHP is a very forgiving programming language, I'd guess. I'll illustrate with code of what I think is going on:

$foo = 'bar'; // $foo is now equal to "bar"  $foo['bar'] = 'foo'; // $foo['bar'] doesn't exists - use first index instead (0) // $foo['bar'] is equal to using $foo[0] // $foo['bar'] points to a character so the string "foo" won't fit // $foo['bar'] will instead be set to the first index // of the string/array "foo", i.e 'f'  echo $foo['bar']; // output will be "f"  echo $foo; // output will be "far"  echo $foo['bar']['bar']; // $foo['bar'][0] is equal calling to $foo['bar']['bar'] // $foo['bar'] points to a character // characters can not be represented as an array, // so we cannot reach anything at position 0 of a character // --> fatal error 

回答2:

I was fighting a similar problem, so documenting here in case useful.

In a __get() method I was using the given argument as a property, as in (simplified example):

function __get($prop) {      return $this->$prop; } 

...i.e. $obj->fred would access the private/protected fred property of the class.

I found that when I needed to reference an array structure within this property it generated the Cannot use String offset as array error. Here's what I did wrong and how to correct it:

function __get($prop) {      // this is wrong, generates the error      return $this->$prop['some key'][0];       // this is correct      $ref = & $this->$prop;      return $prop['some key'][0]; } 

Explanation: in the wrong example, php is interpreting ['some key'] as a key to $prop (a string), whereas we need it to dereference $prop in place. In Perl you could do this by specifying with {} but I don't think this is possible in PHP.

回答3:

I just want to explain my solving for the same problem.

my code before(given same error):

$arr2= ""; // this is the problem and solve by replace this $arr2 = array(); for($i=2;$i"; //and it's($arr2[$i][$j]) give an error: Cannot use string offset as an array     }     $td .=""; } 

my code after and solved it:

$arr2= array(); //change this from $arr2=""; for($i=2;$i";     }     $td .=""; } 

Thank's. Hope it's helped, and sorry if my english mess like boy's room :D

回答4:

When you directly print print_r(($value['']->)); then it shows this fatal error Cannot use string offset as an object in. If you print like this

$var = $value['#node']->; print_r($var);

You won't get the error!!

回答5:

I believe what are you asking about is a variable interpolation in PHP.

Let's do a simple fixture:

$obj = (object) array('foo' => array('bar'), 'property' => 'value'); $var = 'foo'; 

Now we have an object, where:

print_r($obj); 

Will give output:

stdClass Object     (         [foo] => Array             (                 [0] => bar             )          [property] => value     ) 

And we have variable $var containing string "foo".

If you'll try to use:

$give_me_foo = $obj->$var[0]; 

Instead of:

$give_me_foo = $obj->foo[0]; 

You get "Cannot use string offset as an array [...]" error message as a result, because what you are trying to do, is in fact sending message $var[0] to object $obj. And - as you can see from fixture - there is no content of $var[0] defined. Variable $var is a string and not an array.

What you can do in this case is to use curly braces, which will assure that at first is called content of $var, and subsequently the rest of message-sent:

$give_me_foo = $obj->{$var}[0]; 

The result is "bar", as you would expect.

回答6:

I was able to reproduce this once I upgraded to PHP 7. It breaks when you try to force array elements into a string.

foreach ($foo) {   $params = '';   $index = 0;   $params[$index]['keyName'] = $name . '.' . $fileExt; } 

After changing:

$params = ''; 

to:

$params = array(); 

I stopped getting the error. I found the solution in this bug report thread. I hope this helps.

回答7:

The error occurs when:

$a = array();  $a['text1'] = array(); $a['text1']['text2'] = 'sometext'; 

Then

echo $a['text1']['text2'];     //Error!! 

Solution

$b = $a['text1']; echo $b['text2'];    // prints: sometext 

..

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