Question

无论何时我在phpmyadmin中尝试以下查询，都可以正常工作：

SELECT CAST(DATETIME AS DATE), Provider, COUNT(Provider) 
FROM purchases 
GROUP BY CAST(DATETIME AS DATE), Provider 
ORDER BY CAST(DATETIME AS DATE) DESC

尽管如此，当我将其放入PHP脚本时，我总是收到一个布尔值：

$sql="SELECT CAST(DATETIME AS DATE), Provider, COUNT(Provider) 
FROM purchases 
GROUP BY CAST(DATETIME AS DATE), Provider 
ORDER BY CAST(DATETIME AS DATE) DESC";


if (mysqli_query($con,$sql))
  {

    $result = mysql_query($sql);
    while($row = mysql_fetch_assoc($result)) {
    echo $row['CAST(DATETIME AS DATE)'];
    echo $row['Provider'];
    echo $row['COUNT(Provider)'];

}

}
else
  {
  echo "Error creating table: " . mysqli_error($con);
  }

mysqli_close($con)

;

我非常感谢有关如何解决此问题的一些建议。

Answer 1

此代码如下所示，因此可以正常工作

您在所有查询中都使用mysqli，并检查$result是否比打印记录有价值，否则打印了error消息

$result = mysqli_query($con,$sql);
if($result)
{
    while($row = mysqli_fetch_assoc($result)){
        echo $row['CAST(DATETIME AS DATE)'];
        echo $row['Provider'];
        echo $row['COUNT(Provider)'];

    }
}
else
{
    echo "Error creating table: " . mysqli_error($con);
}

mysqli_close($con)

SQL查询工作在phpmyadmin中，但不在PHP脚本中

1 个答案: