PHP字符串嵌套/多维数组 [英] PHP string to nested / multidimensional array
问题描述
我有这样的例子PHP字符串：
$字符串=@ [ITEM_1] [门] @ [莫扎特] [草] = YES @ [莫扎特] [绿] =无@ [莫扎特] [人] @ [蓝] [电影] = YES @ ITEM_1] [节拍] = YES @ [ITEM_1] [音乐] =无
现在$字符串idented只是易观：
- @ [ITEM_1] [门]结果
- @ [莫扎特] [草] = YES结果
- @ [莫扎特] [绿] =无结果
- @ [莫扎特] [人]搜索结果
- @ [蓝] [电影] = YES搜索结果
- @ [ITEM_1] [节拍] = YES结果
- @ [ITEM_1] [音乐] =无结果
我想知道我怎么能得到这个字符串（或以下这种风格等字符串），并在一个数组，看起来像变换：
阵列
（
[ITEM_1] =＆GT;排列
（
[门] =＆GT;排列
（
[莫扎特] =＆GT;排列
（
[草] =＆GT;是
[绿色] =＆GT;没有
[人] =＆GT;排列
（
[蓝] =＆GT;排列
（
[电影] =＆GT;是
）
）
）
）
[拍] =＆GT;是
[音乐] =＆GT;没有
）
）我想什么
我试图使用和递归函数创建一个嵌套数组，但我不能在递归函数访问数组指针（深层次）..不知道为什么..也许是错误的补丁，回答。
谢谢你，解决方案好吧，我希望你还是需要这个，因为我浪费了比我更想联系得到这个正确的时间：）
基本上，我的做法是首先操纵[的]串入格式[设置] [键] =值，然后通过钥匙串回路，并与最后一组按键比较他们创造正确的密钥层次结构。我用的eval因为它更容易，但你可以写一个替换功能，如果你不能忍受看到在code，它的功能：
//首先我们得到的字符串转换成更易于使用CHUNKS WORK
$ original_string =@ [ITEM_1] [门] @ [莫扎特] [草] = YES @ [莫扎特] [绿] =无@ [莫扎特] [人] @ [蓝] [电影] = YES @ [ITEM_1] [击败] = YES @ [ITEM_1] [音乐] =无;
$ cleaned_string = str_replace函数（'] @ ['，'] ['，$ original_string）;
/ *这导致等于值按键集群：
@ [ITEM_1] [门] [莫扎特] [草] = YES @ [莫扎特] [绿] =无@ [莫扎特] [人] [蓝] [电影] = YES @ [ITEM_1] [击败] = YES @ ITEM_1] [音乐] =无OR（与换行符为清楚起见）：@ [ITEM_1] [门] [莫扎特] [草] = YES
@ [莫扎特] [绿] =无
@ [莫扎特] [人] [蓝] [电影] = YES
@ [ITEM_1] [击败] = YES
@ [ITEM_1] [音乐] =无* ///把它分解成一个数组：
$元素=爆炸（'@'，$ cleaned_string）;//创建一个变量最后一个字符串比较
$ last_keys =;
//而另一个将作为我们的最终数组
$ array_of_arrays =阵列（）;
//现在通过每个[ITEM_1] [门] [莫扎特] [草] = YES，[莫扎特] [绿] =无等循环
的foreach（$元素作为$元素）{
如果（$元素==）{继续;} //跳过第一个空项 //打破串入[0] =组键和[1]的值并因此终止了串
//所以[ITEM_1] [门] [莫扎特] [草] = YES变为[ITEM_1] [门] [莫扎特] [草]，是的
$件=爆炸（=，str_replace函数（阵列（[，]），阵列（['，']），修整（$元件）））;
//现在这组按键比较最后一组密钥，如果它们重叠，将它们合并成一个单一的密钥字符串
$ clean_keys = combine_key_strings（$件[0]，$ last_keys）;
//设置新的密钥字符串值下一个比较
$ last_keys = $ clean_keys;
//和（丑，我知道），我们使用一个eval转换[ITEM_1] [门] [莫扎特] [草] ='是'到正确键控阵列
的eval（\\ $ array_of_arrays$ clean_keys='.trim（$件[1]）。';）;
}//现在转储内容
的print_r（$ array_of_arrays）;
//这个函数COMPA
功能combine_key_strings（$新，老$）{
//获取启动新的字符串键
$ new_keys =爆炸（'] ['，$新）;
$ first_key = $ new_keys [0]']'。 //看它是否出现在最后一个字符串中
$ last_occurance = strrpos（$岁，$ first_key）;
//如果是这样，合并两个字符串创建完整KeyString中数组
如果（is_int（$ last_occurance））{
返回SUBSTR（$老，0，$ last_occurance）$新。
}
返回$新;
}这应该吐出你的正确嵌套数组：
阵列
（
[ITEM_1] =＆GT;排列
（
[门] =＆GT;排列
（
[莫扎特] =＆GT;排列
（
[草] =＆GT;是
[绿色] =＆GT;没有
[人] =＆GT;排列
（
[蓝] =＆GT;排列
（
[电影] =＆GT;是
） ） ） ） [拍] =＆GT;是
[音乐] =＆GT;没有
））晚安！
I have this example php string:
$string = "@[item_1][door] @[mozart][grass] = yes @[mozart][green] = no @[mozart][human] @[blue][movie]=yes @[item_1][beat] = yes @[item_1][music] = no ";
now $string idented just to easy view:
- @[item_1][door]
- @[mozart][grass] = yes
- @[mozart][green] = no
- @[mozart][human]
- @[blue][movie]=yes
- @[item_1][beat] = yes
- @[item_1][music] = no
I want to know how can i get this string ( or other string following this style ) and transform in an array that looks like:
Array ( [item_1] => Array ( [door] => Array ( [mozart] => Array ( [grass] => yes [green] => no [human] => Array ( [blue] => Array ( [movie] => yes ) ) ) ) [beat] => yes [music] => no ) )
What i tried
I tried to use and recursive function to create an nested array but i can't have access to the array pointer ( in deep levels ) in recursive functions.. don't know why.. maybe is the wrong patch to the answer. thank you,
解决方案OK, I hope you still need this, because I wasted more time than I'd like to admin getting this right :)
Basically, my approach was to first manipulate the string into the format [set][of][keys]=value, and then loop through the string of keys and comparing them with the last set of keys to create the correct key hierarchy. I used eval because it's easier, but you can write a replacement function if you can't stomach seeing that function in your code:
//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; }This should spit out your correctly nested array:
Array ( [item_1] => Array ( [door] => Array ( [mozart] => Array ( [grass] => yes [green] => no [human] => Array ( [blue] => Array ( [movie] => yes ) ) ) ) [beat] => yes [music] => no ) )Good night!
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