I saw some prompts like this in the php error log:
[27-Aug-2011 22:26:12] PHP Warning: Cannot use a scalar value as an array in /www/hx/enjoy.php on line 14
[ 27-Aug-2011 22:26:18] PHP Warning: Cannot use a scalar value as an array in /www/hx/enjoy.php on line 14
Check the source program, it probably looks like this:
Copy Code The code is as follows:
$arr_hx = $mem->get('hx');
if(!$arr_hx) {
$arr_hx['a'] = 'b';
$mem-> ;set('hx',$arr_hx);
}
Basically understood, when $mem->get does not get the value, it returns false. At this time, $arr_hx is false, a Boolean value, and then it is returned Used as an array, resulting in such a prompt. In fact, it is also a case where the variable is not defined. Add $arr_hx = array() before the assignment to solve the problem.
After checking, there is the following explanation:
Quote
What you need to pay attention to is the type conversion:
If a variable name (such as a) has been defined as a non-array type, such as integer, then a can be converted to a floating point, string (even object type), but not an array, that is, a[0]=1; is wrong, and PHP will issue a warning like "Cannot use a scalar value as an array". Even if a is defined as a one-dimensional array, it cannot be converted to a high-dimensional array.
The above introduces the solution to the scalar PHP warning Cannot use a scalar value as an array, including scalar content. I hope it will be helpful to friends who are interested in PHP tutorials.
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