Question

我对php很新，我想在mysql列上做两个选择Counts并将它们放入一个数组中。我的代码如下，但这不起作用。

// DB connection
  $link = mysqli_connect( "localhost", "username", "password", "CE_TRANSACTIONS" );

// Query users table
  $sql = "SELECT (SELECT COUNT(STATUS) FROM CE_TRANSACTIONS WHERE STATUS = 0),
  (SELECT COUNT(STATUS) FROM CE_TRANSACTIONS WHERE STATUS = 3);"

// Execute query
  $result = mysqli_query($link,$sql);

// Loop over all result rows
  $result_array = array();

     while($array = mysqli_fetch_assoc($result)) 
     {

        $result_array[] = $array[];
     }

// Write to JSON
   echo json_encode($result_array);

嗨谢谢您的快速回复我也收到以下错误消息 PHP解析错误：语法错误，以下行的意外T_VARIABLE。

  // Execute query
  $result = mysqli_query($link,$sql);

Answer 1

更改

 $result_array[] = $array[];

到

 $result_array = $array;

您也可以通过使用以下查询来提升性能。

SELECT Sum(Case When Status = 0 Then 1 Else 0), 
       Sum(Case When Status = 3 Then 1 Else 0)
FROM   CE_TRANSACTIONS

Answer 2

我相信您的问题来自$array[];

[]说的是＆＃39;访问＆＃39;的值，但您没有访问该阵列的任何索引。

您想要做的是： $result_array[] = $array

Answer 3

PHP Parse error: syntax error, unexpected T_VARIABLE是因为你的行



$sql = "SELECT (SELECT COUNT(STATUS) FROM CE_TRANSACTIONS WHERE STATUS = 0),
  (SELECT COUNT(STATUS) FROM CE_TRANSACTIONS WHERE STATUS = 3);"

$sql = "SELECT (SELECT COUNT(STATUS) FROM CE_TRANSACTIONS WHERE STATUS = 0), (SELECT COUNT(STATUS) FROM CE_TRANSACTIONS WHERE STATUS = 3);"

缺少尾随。

将2个sql值添加到数组中

3 个答案: