PHP Warning: Solution to in_array() expects parameter

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Release: 2023-06-23 06:40:01
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During the development process, we may encounter such an error message: PHP Warning: in_array() expects parameter. This error message will appear when using the in_array() function. It may be caused by incorrect parameter passing of the function. Let’s take a look at the solution to this error message.

First of all, you need to clarify the role of the in_array() function: check whether a value exists in the array. The prototype of this function is: in_array($value, $array), where $value represents the value that needs to be found, and $array represents the array that needs to be found.

Solution 1: Check whether the parameter type is correct

When the parameter is passed in the in_array() function, it needs to be a variable or an array. If what is passed is not a variable or array, the above error message will appear. So, to avoid this error, we need to check if the parameter type is correct.

For example, the following code will cause an error message:

in_array("abc", "aabbcc");
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Because the second parameter is not an array, but a string.

Solution 2: Check whether the parameter is empty

When using the in_array() function, if the second parameter is an empty array, the above error message will appear.

For example, the following code will cause an error message:

$arr = [];
in_array("abc", $arr);
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Because the second parameter is an empty array.

Solution 3: Check whether the parameter is null

One of the reasons why the above error message will appear in the in_array() function is that the first parameter is empty (null).

For example, the following code will cause an error message:

$value = null;
$arr = ["abc", "def"];
in_array($value, $arr);
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Because the first parameter $value is empty.

Solution 4: Check whether the data types of parameters are consistent

When the in_array() function checks whether the value exists in the array, you also need to pay attention to whether the data types are consistent. If the data types are inconsistent, the above error message will appear. For example:

$arr = [0, 1, 2, 3];
in_array("0", $arr);
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Although the values ​​​​of 0 and "0" are the same, the data types are different, which will cause an error message to appear.

Finally, it should be noted that in order to avoid such errors, when using the in_array() function, you should first check whether the passed parameter type and data type are correct, and convert them if necessary. Also, avoid passing empty arrays or empty parameters. This can better ensure the correctness and robustness of the program.

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