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采用包含方括号的方括号,作为非阵列
原标题:take php string which contains square brackets as string not array

I have a weird problem here. I m using an associative array in php (using cakePHP) which has the following form:

$my_array = array(
      data[ a ][ b ] =>  value1 , 
      data[ b ][ c ] =>  value2 , 
      data[ b ][ d ] =>  value3 , 
      data[ e ] => array(
             data[ e1 ][ e2 ]  =>  value3 , 
             data[ e1 ][ e3 ]  =>  value4 ));

第一大问题是:

 data[ e1 ][ e2 ]  =>  value3  and  data[ e1 ][ e3 ]  =>  value4 

就像这样的阵列:

 data[ e1 ]  => array(
     [ e2 ] =>  value3 , 
     [ e3 ] =>  value4 );

我不想将这些内容视为阵列,我希望将它们视为阵列数据的关键和价值。 事实上,我希望各阵列的所有元件——记录和数据[(e)]都作为相应阵列的关键和价值(而不是阵列)。

任何帮助?

P.S 似乎只有在我对面包房作假时才会发生这种情况,如果我不使用面包房,一切似乎都是罚款,而“数据”则来自rel=“nofollow”>。

问题回答

你的法典无效。 我最好的猜测是,它应当这样做:

$my_array = array(
     $data[ a ][ b ] =>  value1 ,
     $data[ b ][ c ] =>  value2 ,
     $data[ b ][ d ] =>  value3 ,
     $data[ e ] => array(
            $data[ e1 ][ e2 ] =>  value3 ,
            $data[ e1 ][ e3 ] =>  value4 ));

请向我们展示数据的内容(例如,使用 印本-r)。





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