所以我只想在满足少数条件时更新行。
1. 如果我已经有一个特定行的值
$val = mysqli_real_escape_string($conn,$_POST['$result']);
$sql = "SELECT * FROM user_video WHERE video_id = '$id' AND user_uid = '".$_SESSION['u_uid'] ."'";
$result = mysqli_query($conn, $sql);
$resultCheck = mysqli_num_rows($result);
if($resultCheck>0){
所以如果我已经有一个我不需要插入新值,只想更新值。
2. 如果保存的值小于后置值
if ($row['done']< $val) {
$sql = "UPDATE user_video SET obejrzane='$val' WHERE video_id = '$id' AND user_uid='".$_SESSION['u_uid'] ."'";
mysqli_query($conn, $sql);
}
当我想插入新数据时,我的代码工作正常。但当我想更新时,每次更新时,即使$\u post['$val']比一行中现有的小。。。有什么合适的帮助吗?!
完整代码
<?php
session_start();
include 'db.php';
$val = mysqli_real_escape_string($conn,$_POST['$result']);
$time = mysqli_real_escape_string($conn,date("Y-m-d H:i:s"));
$id = mysqli_real_escape_string($conn,$_POST['$title']);
$sql = "SELECT * FROM user_video WHERE video_id = '$id' AND user_uid = '".$_SESSION['u_uid'] ."'";
$result = mysqli_query($conn, $sql);
$resultCheck = mysqli_num_rows($result);
if($resultCheck>0){
if ($row['done']< $val) {
$sql = "UPDATE user_video SET obejrzane='$val' WHERE video_id = '$id' AND user_uid='".$_SESSION['u_uid'] ."'";
mysqli_query($conn, $sql);
}
echo $resultCheck;
}
else{
if (!empty($val)) {
$sql = "INSERT INTO user_video (user_uid, video_id, done, created) VALUES ('".$_SESSION['u_uid'] ."', '$id', '$val', '$time');";
if ($conn->query($sql) === TRUE) {
echo "New record created successfully";
}
else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
}
}
$conn->close();
1条答案
按热度按时间63lcw9qa1#
您可能没有发布完整的代码,或者您的php出现了错误消息
undefined index 'done'
因为你从来没有真正访问过SELECT
查询。代码应如下所示: