Solution:

The errorIllegal string offset 'whatever' in... generally means: you're trying to use a string as a full array.

That is actually possible since strings are able to be treated as arrays of single characters in php. So you're thinking the $var is an array with a key, but it's just a string with standard numeric keys, for example:

$fruit_counts = array('apples'=>2, 'oranges'=>5, 'pears'=>0);
echo $fruit_counts['oranges']; // echoes 5
$fruit_counts = "an unexpected string assignment";
echo $fruit_counts['oranges']; // causes illegal string offset error
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You can see this in action here: http://ideone.com/fMhmkR

For those who come to this question trying to translate the vagueness of the error into something to do about it, as I was.

Solution:

You're trying to access astring as if it were an array, with a key that's astring.string will not understand that. In code we can see the problem:

"hello"["hello"];
// PHP Warning:  Illegal string offset 'hello' in php shell code on line 1

"hello"[0];
// No errors.

array("hello" => "val")["hello"];
// No errors. This is *probably* what you wanted.
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In depth

Let's see that error:

Warning: Illegal string offset 'port' in ...

What does it say? It says we're trying to use the string'port' as an offset for a string. Like this:

$a_string = "string";

// This is ok:
echo $a_string[0]; // s
echo $a_string[1]; // t
echo $a_string[2]; // r
// ...

// !! Not good:
echo $a_string['port'];
// !! Warning: Illegal string offset 'port' in ...
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What causes this?

For some reason you expected anarray, but you have astring. Just a mix-up. Maybe your variable was changed, maybe it never was anarray, it's really not important.

What can be done?

If we know we should have anarray, we should do some basic debugging to determine why we don't have anarray. If we don't know if we'll have anarray orstring, things become a bit trickier.

What we can do is all sorts of checking to ensure we don't have notices, warnings or errors with things like and or :

$a_string = "string";
$an_array = array('port' => 'the_port');

if (is_array($a_string) && isset($a_string['port'])) {
    // No problem, we'll never get here.
    echo $a_string['port'];
}

if (is_array($an_array) && isset($an_array['port'])) {
    // Ok!
    echo $an_array['port']; // the_port
}

if (is_array($an_array) && isset($an_array['unset_key'])) {
    // No problem again, we won't enter.
    echo $an_array['unset_key'];
}


// Similar, but with array_key_exists
if (is_array($an_array) && array_key_exists('port', $an_array)) {
    // Ok!
    echo $an_array['port']; // the_port
}
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There are some subtle differences between and . For example, if the value of$array['key'] isnull,isset returnsfalse.array_key_exists will just check that, well, the key exists.

Solution:

Please try this way.... I have tested this code.... It works....

$memcachedConfig = array("host" => "127.0.0.1","port" => "11211");
print_r($memcachedConfig['host']);
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Solution:

There are a lot of great answers here - but I found my issue was quite a bit more simple.

I was trying to run the following command:

$x['name']   = $j['name'];
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and I was getting thisillegal string error on$x['name'] because I hadn't defined the array first. So I put the following line of code in before trying to assign things to$x[]:

$x = array();
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and it worked.

Solution:

A little bit late to the question, but for others who are searching: I got this error by initializing with a wrong value (type):

$varName = '';
$varName["x"] = "test"; // causes: Illegal string offset
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The right way is:

 $varName = array();
 $varName["x"] = "test"; // works
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Solution:

As from PHP 5.4 we need to pass the same datatype value that a function expects. For example:

function testimonial($id); // This function expects $id as an integer
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When invoking this function, if a string value is provided like this:

$id = $array['id']; // $id is of string type
testimonial($id); // illegal offset warning
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This will generate an illegal offset warning because of datatype mismatch. In order to solve this, you can use :

$id = settype($array['id'],"integer"); // $id now contains an integer instead of a string
testimonial($id); // now running smoothly
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Solution:

Before to check the array, do this:

if(!is_array($memcachedConfig))
     $memcachedConfig = array();
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Solution:

In my case i change mysql_fetch_assoc to mysql_fetch_array and solve. It takes 3 days to solve :-( and the other versions of my proyect run with fetch assoc.

Solution:

In my case, I solved it when I changed in function that does sql query after:return json_encode($array) then:return $array

Solution:

It works to me:

Testing Code of mine:

$var2['data'] = array ('a'=>'21','b'=>'32','c'=>'55','d'=>'66','e'=>'77');
foreach($var2 as $result)
{  
    $test = $result['c'];
}
print_r($test);
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Output:55

Check it guys. Thanks

Solution:

just use

$memcachedConfig = array();
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before

 print $memcachedConfig['host'];
 print $memcachedConfig['port'];


 Warning: Illegal string offset 'host' in ....
 Warning: Illegal string offset 'port' in ....
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this is because you never define what is $memcachedConfig, so by default are treated by string not arrays..

Solution:

I solved this problem by using trim() function. the issue was with space.

so lets try

$unit_size = []; //please declare the variable type 
$unit_size = exolode("x", $unit_size);
$width  = trim ($unit_size[1] );
$height = trim ($unit_size[2] );
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I hope this will help you.

Solution:

i think the only reason for this message is because target Array is actually an array like string etc (JSON -> {"host": "127.0.0.1"}) variable

Solution:

For PHP

//Setup Array like so
$memcachedConfig = array(
  "host" => "127.0.0.1",
  "port" => "11211"
);

//Always a good practice to check if empty

if(isset($memcachedConfig['host']) && isset($memcachedConfig['port'])){

    //Some codes

    print_r ($memcachedConfig['host']);
    print_r ($memcachedConfig['port']);

}
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Just make sure to check that the value returning is not empty. So this example was for PHP so find out how to check if an array is empty in other languages.

Solution:

Just incase it helps anyone, I was getting this error because I forgot to unserialize a serialized array. That's definitely something I would check if it applies to your case.

Solution:

It's an old one but in case someone can benefit from this. You will also get this error if your array is empty.

In my case I had:

$buyers_array = array();
$buyers_array = tep_get_buyers_info($this_buyer_id); // returns an array
...
echo $buyers_array['firstname'] . ' ' . $buyers_array['lastname']; 
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which I changed to:

$buyers_array = array();
$buyers_array = tep_get_buyers_info($this_buyer_id); // returns an array
...
if(is_array($buyers_array)) {
   echo $buyers_array['firstname'] . ' ' . $buyers_array['lastname']; 
} else {
   echo 'Buyers id ' . $this_buyer_id . ' not found';
}
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