我有一个点击最喜欢的明星,我想检查数据库,如果记录存在它将删除记录,如果它不存在它将插入记录,什么是问题未插入记录删除记录,如果它存在,否则插入
<?php
include "config.php";
header('Content-Type: application/json');
$landmarkid = $_GET['landmarkid'];
$userid = $_GET['userid'];
try {
$query = mysqli_query($con,"SELECT * from favourite WHERE userid =$userid AND L_ID = $landmarkid");
if(mysqli_num_rows($query) > 0)
{
$q1 = mysqli_query($con,"DELETE from favourite WHERE userid =$userid AND L_ID = $landmarkid");
if($q1){
echo '{"Deleted":"true"}';
}
else {
echo '{"Deleted":"false"}';
}
}
else {
$q2 = mysqli_query($con,"INSERT INTO favourite (userid,L_ID) VALUES ($userid, $landmarkid) ");
if($q2){
echo '{"inserted":"true"}';
}
else {
echo '{"inserted":"false"}';
}
}
} catch (Exception $e) {
echo 'Caught exception: ', $e->getMessage(), "\n";
}
?>
真的吗?让__sure__明白'mysqli_query'和'mysql_query'之间的区别,然后再来。 –
@u_mulder抱歉,现在呢? –
现在你必须明白这一行是什么意思'$ query =($ con,“SELECT * from favorite where userid ='$ userid'AND L_ID ='$ landmarkid'”);' –