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php - 显示查询失败的值

问题描述

这是一个查找 $sql 条件的基本查询。

$sql = "SELECT jobs.pickup as myPickup, jobs.dropoff as myDropoff FROM jobs, locality WHERE 
locality.pickup=jobs.pickup AND locality.dropOff=jobs.dropoff";
$test= mysqli_query($conn, $sql);

  while( $row = mysqli_fetch_assoc($test)) {
echo "Pick up: " . $row["myPickup"]. " to ".$row["myDropoff"]."<br>";

您将如何显示所有失败的值。我试过了

$row != mysqli_fetch_assoc($test)
//and
while($row = mysqli_fetch_assoc($result) == false)

但似乎一无所获。我还尝试通过以下代码编辑 SQL：

SELECT jobs.pickup as myPickup, jobs.dropoff as myDropoff FROM jobs, locality WHERE locality.pickup!=jobs.pickup AND locality.dropOff!=jobs.dropoff

但这也没有用，由于某种原因，SQL“locality.pickup=jobs.pickup AND locality.dropOff=jobs.dropoff”代码中的“AND”语句就像一个“OR”函数一样。 。 诡异的

标签： phpmysql

解决方案

我在 SQL 中尝试了“WHERE NOT”，但我不知道您需要添加括号。这就是为什么它的行为不正确

SELECT jobs.pickup as myPickup, jobs.dropoff as myDropoff FROM jobs, locality WHERE NOT (locality.pickup=jobs.pickup AND locality.dropOff=jobs.dropoff);

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