Parameter usage of json_decode and var_export in php_PHP tutorial

This article mainly talks about the usage of the second parameter of json_decode and var_export in php. Friends who need to know the usage of json_decode and var_export in php can also refer to it.

Both json_decode and var_export have a second parameter. I have never noticed it before, T_T. Moreover, this is the first time I know about this var_export function. The foundation is still not solid.

json_decode
(PHP 5 >= 5.2.0, PECL json:1.2.0-1.2.1)
json_decode — Decode a JSON-formatted string
Description
mixed json_decode ( string $json [, bool $assoc ] )
Accepts a JSON formatted string and converts it into a PHP variable

In the past, json was used as a data format for ajax processing. Today, the data format returned by the interface provided by the partner is json. At that time, I was still thinking that this thing returns object, which can be directly used in javascript. But how to convert this object into an array in php? After scratching my head for a long time, I decided to read the manual. You don’t know it by looking at it, but you are shocked when you see it. It turns out that this thing has a second parameter. If it is passed as true, the decoded data will become an array. Happy.

When array is a continuous array starting from 0, the result of json_encode is a string enclosed by []

When the array is an array that does not start from 0 or is not continuous, the result of json_encode is a string in key-value pattern enclosed by {}

The code is as follows

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代码如下	复制代码
$test = array(); $test[] = 1; $test[] = 1; $test[] = 1; DEBUG(json_encode($test));结果： [1,1,1] $test = array(); $test[] = 1; $test[] = 1; $test[] = 1; unset($test[0]); DEBUG(json_encode($test));

$test = array();

$test[] = 1;

$test[] = 1;

$test[] = 1;

DEBUG(json_encode($test)); Result:

[1,1,1]

$test = array();

$test[] = 1;

$test[] = 1;

$test[] = 1;
unset($test[0]);
DEBUG(json_encode($test));

Result:

{"1":1,"2":1}2. When the string is in the pattern [1,1,1], the result parsed by json_decode is an array by default,

When the string is in the pattern {"1":1,"2":1}, the result parsed by json_decode is an object by default. At this time, you can set its second parameter to true to force it to Return array

3. The above situation occurs because PHP cannot distinguish between one-dimensional arrays and two-dimensional arrays, because it is recommended to set the second parameter to true when using json encoding

代码如下	复制代码
$handle = fopen($file_name, 'w+'); fwrite($handle,'"); fclose($handle);

-------------------------------------------------- ---------------------------------- I want to store the reconstructed data in a file. The data format is an array. The previous method was serialize, and then deserialized when it was taken out. I was lazy today and wondered if I could directly store the array in the file? But the format of arrays is really difficult to spell, especially associative arrays. If I knew all about subscripts, wouldn’t it exhaust me to death? Let’s ask Google. var_export (PHP 4 >= 4.2.0, PHP 5) var_export — Export or return a string representation of a variable Description mixed var_export ( mixed $expression [, bool $return ] ) This function returns structural information about the variable passed to the function. It is similar to var_dump(), except that the representation returned is legal PHP code. You can return a representation of a variable by setting the function's second argument to TRUE. Did you see that? This function is so cool. I have always used var_dump before. I didn’t know someone named var had such a thing. It’s good. It’s good.

The code is as follows	Copy code
$handle = fopen($file_name, 'w+'); fwrite($handle,'"); fclose($handle);

When I use the result, there is still a problem. The function outputs the contents of the array to the page, but there is nothing in the file. Isn’t this troublesome? If that's all, why should I use it? It is displayed on the page, I use

 how cool it is. I was depressed, but after a closer look, it turned out that this function also has a second parameter, which has the same function as json_decode. It seems that my carelessness is still very serious. 

 

	  

		
 代码如下
		
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$handle = fopen($file_name, 'w+');

fwrite($handle,'");

fclose($handle);
	  
	


Examples


 The code is as follows

	  

		
 代码如下
		
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$res = yblog_mspconfiginit("ratings");

var_dump($res);

var_export($res);
/*结果:resource(1) of type (yahoo_yblog)NULL*/

	  
	
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$res = yblog_mspconfiginit("ratings");
var_dump($res);

	  

		
 代码如下
		
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$res = fopen('status.html', 'r');

var_dump($res);

var_export($res);
/*结果:resource(2) of type (stream)NULL*/

	  
	
var_export($res);

/*Result:resource(1) of type (yahoo_yblog)NULL*/


 
Another example:



 The code is as follows


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$res = fopen('status.html', 'r');
var_dump($res);
var_export($res);
/*Result:resource(2) of type (stream)NULL*/



var_export must return legal PHP code. In other words, the code returned by var_export can be directly assigned to a variable as PHP code. And this variable will get the same type of value as var_export
However, when the variable type is resource, it cannot be copied simply. Therefore, when the variable of var_export is of resource type, var_export will return NULL


http://www.bkjia.com/PHPjc/632205.htmlwww.bkjia.comtruehttp: //www.bkjia.com/PHPjc/632205.htmlTechArticleThis article mainly talks about the usage of the second parameter of json_decode and var_export in php. If you need to know about php Friends who know the usage of json_decode and var_export can also refer to it. No matter...