显示查询失败的值

Question

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>";

你将如何展示所有失败的价值观。我试过

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

但似乎没有进展。我还试着用以下代码编辑SQL：

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

但这也不起作用，出于某种原因；"与"；SQL中的语句"；locality.pickup=jobs.pickup AND locality.dropOff=jobs.dropOff"；代码就像一个"；OR"；作用奇怪的

Answer 1

我尝试了"其中NOT"；在SQL中，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);