我想我的Android应用程序连接到一个在线数据库,但SQL错误的过程中弹出错误的SQL语法在PHP脚本
失败您的SQL语法错误;检查 对应于您MariaDB的服务器版本,在1号线
此使用 附近“QRY”正确的语法本手册是PHP脚本
<?php
$name = "localhost";
$username ="**********";
$password ="**********";
$servername ="*********";
$conn = mysqli_connect($name,$username,$password,$servername);
$typee=$POST["typee"];
$heighte=$POST["heighte"];
$weighte=$POST["weighte"];
$ranke=$POST["ranke"];
$typee = $conn->real_escape_string($typee);
$hieghte = $conn->real_escape_string($hieghte);
$weighte = $conn->real_escape_string($weighte);
$ranke = $conn->real_escape_string($ranke);
$qry = "INSERT INTO Naruto(typee,heighte,weighte,ranke)VALUES('$typee','$heighte','$weighte','$ranke')";
if(mysqli_query($conn,qry)===TRUE)
echo "Success";
else
echo "Failed ".$conn->error;
$conn->close();
不要混用mysqli_面向对象和mysqli_过程样式。 'mysqli_query($ conn,qry)'应该是'$ conn-> query($ qry)'。改变它然后回报。编辑:哈,还有,你在'qry'之前缺少'$' – Terminus
谢谢。有效!! –