我厌倦了这个错误。我很确定它可以与“$_POST[name]”一起使用,但 sql 不接受它。
它给我错误提示:您的 SQL 语法有错误;检查与您的 MySQL 服务器版本相对应的手册,了解在第 2 行“)”附近使用的正确语法
$syn = mysql_real_escape_string($_POST['syn']);
$fore = mysql_real_escape_string($_POST['fore']);
$localfore = mysql_real_escape_string($_POST['localfore']);
$save = mysql_query("INSERT INTO tblforecast (Issued,Valid,Synopsis,Forecast,Local_Forecast,Station11,Station12,Station13,Station14,Station15,Station16,Station17,Station18,Station19,Forecaster)
VALUES (now(),'24','$syn','$fore','$localfore','sample','$sample','sample','sample','sample','sample','sample','sample','sample',$id)");
发生什么事了?
PS。第 2 行位于 VALUES 的开头最佳答案
试试这个
$sql = "INSERT INTO tblforecast (Issued,Valid,Synopsis,Forecast,Local_Forecast,Station11,Station12,Station13,Station14,Station15,Station16,Station17,Station18,Station19,Forecaster)
VALUES ('{$datetime}',24,'{$_POST[syn]}','{$_POST[fore]}','{$_POST[localfore]}','sample','sample','sample','sample','sample','sample','sample','sample','sample',$id)";
echo sql;
$save = mysql_query($sql);
关于php - Mysql 查询在 SQL 语法正确时显示错误,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/20899047/