PHP array_intersect或in_array那么MySQL [英] PHP array_intersect or in_array then MYSQL

问题描述

需要找到，如果两个数组匹配，然后在那里他们从他们匹配的mysql匹配一行提取数据。我应该使用

  $ SQL =SELECT * FROM围绕;
 $ resultsd = $ conn-＆GT;查询（$的SQL）;
 的foreach（$ resultsd如rowd $）{} 如果（array_intersect（$ AR1，$ AR2））{
    $剑[] = $ rowd ['TIM'];
    }
 

或者我应该用

 如果（in_array（$ AR1，AR2 $））{
       $剑[] = $ rowd ['TIM'];
   }
 

获取阵列，如：

  $ AR1 [] = $ rowd [净息差'];
   $ AR2 [] = $ rowd [净息差'];
 

那么一个人如何去拉动特定行，他们在匹配？

我看到它们匹配，并打印出数组的好：

 阵列（[0] =＆GT; dcbabcbded）阵列（[0] =＆GT; fafeafaebee [1] =＆GT; afabfdefcbb [2] =＆GT; dcbabcbded
 

但是，当我尝试了回音，他们符合我失败了MySQL的数据：

 阵列（）
 

解决方案

使用 $新= array_intersect（阵列（[0] =＆GT; dcbabcbded），阵列（[0] =＆GT; fafeafaebee [ 1] =＆GT; afabfdefcbb [2] =＆GT; dcbabcbded））：然后进行有效的SQL如上面我的回答说。

如果妳确保你已经拥有百达阵intersenct有一个或没有值，则使SQL只是一个数组元素： $ AR [0] 使用SQL WHERE 子句。

Need to find if two arrays match, and then where they match pull data from the mysql row in which they match. Should I use

 $sql = "SELECT * FROM around";
 $resultsd = $conn->query($sql);
 foreach($resultsd as $rowd) {}

 if (array_intersect($ar1, $ar2)) {
    $sword[] = $rowd['TIM'];
    }

or should I use

 if (in_array($ar1, $ar2)) {
       $sword[] = $rowd['TIM'];
   }

Getting the arrays like:

   $ar1[] = $rowd['nim']; 
   $ar2[] = $rowd['nim']; 

Then how does one go about pulling the specific row that they match at?

I am seeing that they match, and printing out the array's okay:

Array ( [0] => dcbabcbded ) Array ( [0] => fafeafaebee [1] => afabfdefcbb [2] => dcbabcbded

But when I trying a echo the mysql data where they match I fail:

 Array ( )  

解决方案

use $new = array_intersect(Array ( [0] => dcbabcbded ), Array ( [0] => fafeafaebee [1] => afabfdefcbb [2] => dcbabcbded)) : then make valid sql as stated in my answer above.

If u are sure you've allways have array intersenct with one or none values, then make sql just with first array element: $ar[0] using sql WHERE clause.

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