我有这个示例 php 字符串:
$string = "@[item_1][door] @[mozart][grass] = yes @[mozart][green] = no @[mozart][human] @[blue][movie]=yes @[item_1][beat] = yes @[item_1][music] = no ";
现在识别 $string 只是为了便于查看:
- @[item_1][door]
- @[mozart][grass] = yes
- @[mozart][green] = no
- @[mozart][human]
- @[blue][movie]=yes
- @[item_1][beat] = yes
- @[item_1][music] = no
我想知道如何获取此字符串(或遵循此样式的其他字符串)并转换成如下所示的数组:
Array
(
[item_1] => Array
(
[door] => Array
(
[mozart] => Array
(
[grass] => yes
[green] => no
[human] => Array
(
[blue] => Array
(
[movie] => yes
)
)
)
)
[beat] => yes
[music] => no
)
)
我尝试了什么
我尝试使用递归函数来创建嵌套数组,但我无法访问递归函数中的数组指针(深层)。不知道为什么。可能是错误的补丁回答。 谢谢,
最佳答案
好的,我希望你仍然需要这个,因为我浪费的时间比我想管理的时间还多:)
基本上,我的方法是先将字符串操作成 [set][of][keys]=value 的格式,然后循环遍历键字符串并将它们与最后一组键进行比较以创建正确的键等级制度。我使用 eval 是因为它更简单,但如果您无法忍受在代码中看到该函数,您可以编写替换函数:
//FIRST WE GET THE STRING INTO EASIER TO WORK WITH CHUNKS
$original_string = "@[item_1][door] @[mozart][grass] = yes @[mozart][green] = no @[mozart][human] @[blue][movie]=yes @[item_1][beat] = yes @[item_1][music] = no ";
$cleaned_string = str_replace('] @[','][',$original_string);
/* This results in clusters of keys that equal a value:
@[item_1][door][mozart][grass] = yes @[mozart][green] = no @[mozart][human][blue][movie]=yes @[item_1][beat] = yes @[item_1][music] = no
OR (with line breaks for clarity):
@[item_1][door][mozart][grass] = yes
@[mozart][green] = no
@[mozart][human][blue][movie]=yes
@[item_1][beat] = yes
@[item_1][music] = no */
//break it up into an array:
$elements = explode('@',$cleaned_string);
//create a variable to compare the last string to
$last_keys = "";
//and another that will serve as our final array
$array_of_arrays = array();
//now loop through each [item_1][door][mozart][grass] = yes,[mozart][green] = no, etc
foreach($elements as $element){
if ($element==""){continue;} //skip the first empty item
//break the string into [0] = group of keys and [1] the value that terminates the string
//so [item_1][door][mozart][grass] = yes BECOMES [item_1][door][mozart][grass], AND yes
$pieces = explode('=',str_replace(array('[',']'),array("['","']"),trim($element)));
//now compare this set of keys to the last set of keys, and if they overlap merge them into a single key string
$clean_keys = combine_key_strings($pieces[0],$last_keys);
//set the new key string the value for the next comparison
$last_keys = $clean_keys;
//and (ugly, I know) we use an eval to convert "[item_1][door][mozart][grass]='yes'" into a properly keyed array
eval("\$array_of_arrays".$clean_keys." = '".trim($pieces[1])."';");
}
//now dump the contents
print_r($array_of_arrays);
//THIS FUNCTION COMPA
function combine_key_strings($new,$old){
//get the key that starts the newer string
$new_keys = explode('][',$new);
$first_key = $new_keys[0].']';
//see if it appears in the last string
$last_occurance = strrpos ($old,$first_key);
//if so, merge the two strings to create the full array keystring
if (is_int($last_occurance)){
return substr($old,0,$last_occurance).$new;
}
return $new;
}
这应该吐出正确嵌套的数组:
Array
(
[item_1] => Array
(
[door] => Array
(
[mozart] => Array
(
[grass] => yes
[green] => no
[human] => Array
(
[blue] => Array
(
[movie] => yes
)
)
)
)
[beat] => yes
[music] => no
)
)
晚安!
关于PHP 字符串到嵌套/多维数组,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/8676339/