How to convert json into array in php

In web development, JSON (JavaScript Object Notation) is a commonly used data exchange format. In PHP, you can easily convert JSON formatted data into a PHP array using the json_decode() function.

Use the json_decode() function to convert JSON into a PHP array

Use the json_decode() function in PHP to convert JSON format data into an array, the syntax of this function For:

mixed json_decode ( string $json [, bool $assoc = false [, int $depth = 512 [, int $options = 0 ]]] )

Parameter description:

1. json: JSON string to be decoded.
2. assoc: When this parameter is true, an object in array form will be returned instead of the default object form.
3. depth: Specify the maximum parsing depth. The default value is 512.
4. options: Options for controlling parsing behavior, including JSON_BIGINT_AS_STRING, JSON_INVALID_UTF8_IGNORE, JSON_INVALID_UTF8_SUBSTITUTE, JSON_OBJECT_AS_ARRAY, JSON_THROW_ON_ERROR. The default value is 0.

The following is a sample code that converts JSON into a PHP array:

<?php
// JSON字符串
$json_str = &#39;{"name":"Tom", "age":18, "gender":"male"}&#39;;

// 将JSON字符串转换成PHP数组
$arr = json_decode($json_str, true);

// 输出转换后的PHP数组
print_r($arr);
?>

The output result is:

Array
(
    [name] => Tom
    [age] => 18
    [gender] => male
)

In the above code, first a JSON string, then use the json_decode() function to convert the JSON string into a PHP array, and finally use the print_r() function to output the converted PHP array.

If you do not need to convert JSON to PHP array form, but keep it as JSON object form, you can omit the second parameter in the json_decode() function. The sample code is as follows:

<?php
// JSON字符串
$json_str = &#39;{"name":"Tom", "age":18, "gender":"male"}&#39;;

// 将JSON字符串转换成JSON对象
$obj = json_decode($json_str);

// 输出转换后的JSON对象
print_r($obj);
?>

The output result is:

stdClass Object
(
    [name] => Tom
    [age] => 18
    [gender] => male
)

In the above code, the second parameter is not passed in the json_decode() function, so after conversion The result is in the form of a JSON object, not an array.

Use the json_last_error() function to check JSON decoding errors

When using the json_decode() function to decode JSON, if the JSON string format is incorrect, the decoding process may An error occurs. At this time, you can use the json_last_error() function to check JSON decoding errors. The syntax of this function is:

int json_last_error ( void )

Here is a sample code to check for JSON decoding errors:

<?php
// JSON字符串
$json_str = &#39;{"name":"Tom", "age":"18", "gender":"male"}&#39;;

// 将JSON字符串转换成PHP数组
$arr = json_decode($json_str, true);

// 检查解码是否出错
if (json_last_error() != JSON_ERROR_NONE) {
    echo &#39;JSON decode error: &#39; . json_last_error_msg();
} else {
    print_r($arr);
}
?>

In the above code, since the value of the age attribute is a string instead of a number, so the JSON string cannot be converted correctly into a PHP array. During the decoding process, it was found that there was a JSON decoding error, so the error message was output through the json_last_error_msg() function. The output result is:

JSON decode error: Malformed UTF-8 characters, possibly incorrectly encoded

Summary

This article introduces how to use the json_decode() function in PHP to convert JSON format data into a PHP array, and how to Use the json_last_error() function to check for JSON decoding errors. In actual development, we often need to use JSON format data for data exchange, so it is very important for web developers to be proficient in converting JSON into PHP arrays.

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