PHP变量不显示值

时间:2013-10-09 10:56:23

标签: php mysql sql select

我将MySQL Query值存储在PHP变量中,但它不显示数据。 P.S:数据在MySQL表格列中可用。

 <?php 
$cmsca= mysql_query("SELECT SUM(qa_effort) FROM tbl_uat WHERE product='CAP'");
     while ($cresulta = mysql_fetch_array ($cmsca)) 
         $arra[0] = $cresulta[0];
         echo $arra[0];
?>

我没有线索,上面的代码有什么问题?需要帮助!

此致

3 个答案:

答案 0 :(得分:5)

试试这个

   <?php 
  $cmsca= mysql_query("SELECT SUM(qa_effort) as sums FROM tbl_uat WHERE product='CAP'");
   while ($cresulta = mysql_fetch_array($cmsca)) 
   {
     echo $cresulta['sums'];
   }
  ?>

答案 1 :(得分:-1)

首先,不要使用mysql_query, - 不推荐使用,而是使用http://www.php.net/manual/en/mysqli.query.php。 接下来,在运行查询之前,您需要连接到db;

$mysqli = new mysqli("localhost", "my_user", "my_password", "world");

/* check connection */
if ($mysqli->connect_errno) {
    printf("Connect failed: %s\n", $mysqli->connect_error);
    exit();
}

$result = mysqli->query("SELECT SUM(qa_effort) as sums FROM tbl_uat WHERE product='CAP'");
while ($row = $result->fetch_array()) {
    var_dump($row);
}

$mysqli->close();
?>

答案 2 :(得分:-1)

尝试这个怎么样:

<?php
 $arra = array(); 
 $cmsca= mysql_query("SELECT SUM(qa_effort) FROM tbl_uat WHERE product='CAP'");
 while ($row = mysql_fetch_array ($cmsca)) 
     $arra = $row;
     print_r($arra);
 ?>