Question

错误：您的SQL语法出错;查看与您的MySQL服务器版本对应的手册，以便在第1行的“1”附近使用正确的语法。代码如下所示我得到错误，这是从http://cms2.br-de.tk/editinfo.php到{{3} }

<?php
    mysql_connect("mysql10.000webhost.com","******_12","*******") or       die("Error:".mysql_error()); 
    mysql_select_db("******_1");//add your dbname

    //get the variables we transmitted from the form
    $Title = $_POST['Title'];
    $Author = $_POST['Author'];
    $Date = $_POST['Date'];
    $Content = $_POST['Content'];

    //replace TestTable with the name of your table
    //replace id with the ID of your user
    $sql = "UPDATE `posts` SET `Tilte` = '$Tilte',`Author` = '$Author',`Date` =   '$Date',`Content` = '$Content' WHERE `posts`.`ID` = '$ID' 1 ";

     mysql_query($sql) or die ("Error: ".mysql_error());

    echo "Database updated. <a href='editinfo.php'>Return to edit info</a>";

?>

Answer 1

您在查询结尾处添加了额外的1。它应该是这样的：

$sql = "UPDATE `posts` SET `Tilte` = '$Title',`Author` = '$Author',`Date` =   '$Date',`Content` = '$Content' WHERE `posts`.`ID` = '$ID'";

Answer 2

您的陈述结尾处有一个备用1。

UPDATE `posts` SET `Tilte` = '$Title',`Author` = '$Author',`Date` =   '$Date',`Content` = '$Content' WHERE `posts`.`ID` = '$ID';"

正如Grigore正确发现的那样，根据您的列名，您的语句中也可能会出现拼写错误。

UPDATE `posts` SET `Title` = '$Title',`Author` = '$Author',`Date` =   '$Date',`Content` = '$Content' WHERE `posts`.`ID` = '$ID';"

Answer 3

`Tilte` = '$Title'

也许这是标题而不是倾斜，除了在查询结尾处有一个“1”

错误：SQL语法中有错误;检查对应的手册..在第1行“1”附近使用的语法

3 个答案: