PHP MySQL后期插入

我在使用php脚本向mysql数据库插入值时遇到问题

<?php

$conn = mysqli_connect($dbhost, $dbuser, $dbpass, $db) or die (mysqli_error($conn)); mysqli_select_db($conn, $db) or die (mysqli_error($conn));

$token = null; $lastWaterDate = null; $lastWaterAmount = null; $zipCode = null; $waterToDate = null; $addKey = null;

if(isset ($_POST["token"])){
    $token = $_POST["token"]; 
} 
if(isset ($_POST["lastWaterDate"])){
    $lastWaterDate = $_POST["lastWaterDate"]; 
} 
if(isset ($_POST["lastWaterAmount"])){
    $lastWaterAmount = $_POST["lastWaterAmount"]; 
} 
if(isset ($_POST["zipCode"])){
    $zipCode = $_POST["zipCode"]; 
} 
if(isset ($_POST["waterToDate"])){
    $waterToDate = $_POST["waterToDate"]; 
} 
if(isset ($_POST["addKey"])){
    $addKey =$_POST["addKey"]; 
}

if($addKey == -1){
    $sql = "INSERT INTO UserInfo (userToken) VALUES ('$token')"; 
}

elseif($addKey == 0){
    $sql = "INSERT INTO UserInfo (userToken, lastWaterDate, lastWaterAmount, zipCode, waterToDate) VALUES ('$token','$lastWaterDate','$lastWaterAmount','$zipCode','$waterToDate') ON DUPLICATE KEY UPDATE lastWaterDate=VALUES(lastWaterDate), lastWaterAmount=VALUES(lastWaterAmount), waterToDate=VALUES(waterToDate)"; 
}

else{
    $sql = "INSERT INTO UserInfo (userToken, lastWaterDate, lastWaterAmount, zipCode, waterToDate) VALUES ('$token','$lastWaterDate','$lastWaterAmount','$zipCode','$waterToDate') ON DUPLICATE KEY UPDATE lastWaterDate=VALUES(lastWaterDate), lastWaterAmount=VALUES(lastWaterAmount), zipCode=VALUES(zipCode), waterToDate=VALUES(waterToDate)"; 
}

$res = mysqli_query($conn, $sql) or die(mysqli_error($conn));

if($res){
    echo "Successfully updated db"; 
} 
else{
    echo "Failed to update db"; 
}

mysqli_close($conn);

?>

---

当addkey不是-1时，是否要更新表，而不是插入

有关更新和插入的差异的更多信息，请查看此项：

另外，你能提供一些关于你的数据库和你得到的错误的更多信息吗？（如果你得到任何）

您的问题很可能与实现$sql查询的方式有关。关注脚本的这一部分，问题可能是您从表单接收到的字符串是零“0”，而不是整数。提交表单时，您的条件应为整数0。您需要将其强制为int，然后在条件上使用$\u POST。请始终转义post变量

<?php
$conn = mysqli_connect('localhost', 'xxx', 'xxx', 'xxx') or die (mysqli_error($conn));

mysqli_select_db($conn, 'test') or die (mysqli_error($conn));

$_POST['token'] = 'abca';
$_POST['lastWaterDate'] = '2017-12-12';
$_POST['lastWaterAmount'] = '$100';
$_POST['zipCode'] = '01010';
$_POST['waterToDate'] = '';

// Let's ensure that the value is an INT
$_POST['addKey'] = (int)'0';

if(isset($_POST["token"])){
    $token = mysqli_real_escape_string($conn, $_POST["token"]);
}

if(isset($_POST["lastWaterDate"])){
    $lastWaterDate = mysqli_real_escape_string($conn, $_POST["lastWaterDate"]);
}

if(isset($_POST["lastWaterAmount"])){
    $lastWaterAmount = mysqli_real_escape_string($conn, $_POST["lastWaterAmount"]);
}

if(isset($_POST["zipCode"])){
    $zipCode = mysqli_real_escape_string($conn, $_POST["zipCode"]);
}

if(isset($_POST["waterToDate"])){
    $waterToDate = mysqli_real_escape_string($conn, $_POST["waterToDate"]);
}

if(isset($_POST["addKey"])){
    $addKey = mysqli_real_escape_string($conn, $_POST["addKey"]);
}

if($_POST['addKey'] == -1){
    $sql = "INSERT INTO UserInfo (userToken) VALUES ('$token')";
} elseif($_POST['addKey'] == 0) {
   $sql = "INSERT INTO UserInfo (userToken, lastWaterDate, lastWaterAmount, zipCode, waterToDate) VALUES ('$token','$lastWaterDate','$lastWaterAmount','$zipCode','$waterToDate') ON DUPLICATE KEY UPDATE lastWaterDate=VALUES(lastWaterDate), lastWaterAmount=VALUES(lastWaterAmount), waterToDate=VALUES(waterToDate)";
}else{
    $sql = "INSERT INTO UserInfo (userToken, lastWaterDate, lastWaterAmount, zipCode, waterToDate) VALUES ('$token','$lastWaterDate','$lastWaterAmount','$zipCode','$waterToDate') ON DUPLICATE KEY UPDATE lastWaterDate=VALUES(lastWaterDate), lastWaterAmount=VALUES(lastWaterAmount), zipCode=VALUES(zipCode), waterToDate=VALUES(waterToDate)";
}

$res = mysqli_query($conn, $sql);

if($res){
    echo "Successfully updated db";
}
else{
    echo "Failed to update db";
}

如果您
回显$sql
然后手动执行该查询，会发生什么情况？您是否收到消息
成功更新了db
？您将永远不会收到消息
更新db失败，因为当它失败时，它会消失（mysqli\u错误（$conn）
。你应该学会使用准备好的语句，而不是替换变量，以防止SQL注入。你说过当addKey=-1时“插入工作正常”。这是否意味着将记录插入数据库？但随后你会说“数据库中似乎什么都没有发生”。到底是什么问题？我尝试了“$SQL=“更新用户信息集lastWaterDate=$lastWaterDate，lastWaterAmount=$lastWaterAmount，zipCode=$zipCode，waterToDate=$waterToDate，其中userToken=$token”；”但这也没有实现help@mooooo098这回答了你的问题吗？我的回答对你有帮助吗？如果有，请毫不犹豫地将其标记为你的最佳答案。