我正在尝试通过包含 3 个表的联接查询获取消息 一切都按预期工作正常,但是当我尝试限制结果时,查询失败并且不返回任何内容
查询
SELECT
P.patient_name,
DM.message,
DM.secsince,
DM.id,
DM.pid,
DM.did,
DM.mode,
DM.date,
DM.seen
FROM tbl_patient P,
tbl_doct D,
tbl_doct_messages DM
WHERE (DM.did = D.id AND DM.pid = P.id)
AND (DM.mode = 'in')
AND (D.id = '$id')
ORDER BY DM.secsince DESC;
这个查询工作正常,并给了我预期的结果
但是当我使用 LIMIT 时,它失败并且由于某种原因没有返回任何内容
查询
SELECT
P.patient_name,
DM.message,
DM.secsince,
DM.id,
DM.pid,
DM.did,
DM.mode,
DM.date,
DM.seen
FROM tbl_patient P,
tbl_doct D,
tbl_doct_messages DM
WHERE (DM.did = D.id AND DM.pid = P.id)
AND (DM.mode = 'in')
AND (D.id = '$id')
ORDER BY
DM.secsince DESC;
LIMIT=1;
这是整个 php 代码
<?php
include("conn.php");
if(mysqli_connect_error($con)) {
echo "Failed To Connect";
}
$id = $_GET['id'];
$qry = "SELECT P.patient_name, DM.message, DM.secsince, DM.id, DM.pid, DM.did, DM.mode, DM.date, DM.seen
FROM tbl_patient P, tbl_doct D, tbl_doct_messages DM
WHERE (DM.did = D.id AND DM.pid = P.id) AND (DM.mode = 'in') AND (D.id = '$id')
ORDER BY DM.secsince DESC
LIMIT=1;";
$res = mysqli_query($con, $qry);
$flag = array();
while(($row = mysqli_fetch_array($res))){
array_push($flag, $row);
}
echo json_encode($flag);
mysqli_close($con);
?>
但没有结果,因此当我添加限制时,mysqli_fetch_array 失败
为什么会发生这种情况?我基本上没有主意了。
最佳答案
更新您的查询
来自
SELECT
P.patient_name,
DM.message,
DM.secsince,
DM.id, DM.pid,
DM.did, DM.mode,
DM.date, DM.seen
FROM
tbl_patient P,
tbl_doct D,
tbl_doct_messages DM
WHERE
(DM.did = D.id AND DM.pid = P.id) AND
(DM.mode = 'in') AND (D.id = '$id')
ORDER BY
DM.secsince DESC
LIMIT=1
删除=
SELECT
P.patient_name,
DM.message,
DM.secsince,
DM.id, DM.pid,
DM.did, DM.mode,
DM.date, DM.seen
FROM
tbl_patient P,
tbl_doct D,
tbl_doct_messages DM
WHERE
(DM.did = D.id AND DM.pid = P.id) AND
(DM.mode = 'in') AND (D.id = '$id')
ORDER BY
DM.secsince DESC
LIMIT 1
关于PHP Mysql LIMIT 查询失败,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/55108415/