我遇到错误，无法修复。
response error:
Database query failedYou have an error in your SQL syntax;
check the manual that corresponds to your MySQL server version
for the right syntax to use near '' as Loreta Punzalan'},
gender = {'Female'}, dob = {'01/23/20' at line 2
JS 阿贾克斯
$("button.update_btn").click(function(event){
//alert("test");
var clickedEditid = this.id.split('-');
var dbId = clickedEditid[1];
var mydata = '&recordToBeUpdated='+ dbId;
var mydatas = $('#instructorUps').serialize();
var datas = mydatas+mydata;
//alert(datas);
$.ajax({
type:"POST",
url:"<?php echo get_stylesheet_directory_uri(); ?>/includes/response_instructor_update.php?>",
dataType:"text",
data: datas,
success: function(response){
alert(response);
// $('#ins_'+dbId).fadeIn();
//$('#instructorModal').append(response);
location.reload();
},
error:function(xhr, ajaxOption, thrownError){
alert(thrownError);
}
});
});
Response_instructor_update.php
<?php
// Response for AJAX
require_once("connection.php");
require_once("functions.php");
if(isset($_POST['fullnameup'])){
$idupdate = mysql_prep($_POST["recordToBeUpdated"]);
$fullname = mysql_prep($_POST["fullnameup"]);
$username = mysql_prep($_POST["usernameup"]);
$gender = mysql_prep($_POST["genderup"]);
$dob = mysql_prep($_POST["dobup"]);
$nationality = mysql_prep($_POST["nationalityup"]);
$mobile = mysql_prep($_POST["mobileup"]);
$emailadd = mysql_prep($_POST["emailaddup"]);
$address = mysql_prep($_POST["addressup"]);
$zipcode = mysql_prep($_POST["zipcodeup"]);
$query = "UPDATE instructor SET
instructor = {'$fullname'},
gender = {'$gender'},
dob = {'$dob'},
nationality = {'$nationality'},
mobile = {$mobile},
emailadd = {'$emailadd'},
address = {'$address'},
zipcode = {$zipcode},
username = {'$username'}
WHERE instructorid= {$idupdate}";
$updatedInstructor = mysql_query($query,$connection);
confirm_query($updatedInstructor);
if (mysql_affected_rows() == 1){
echo "TRUE";
mysql_close($connection);
}else{
die("Database query failed". mysql_error());
}
}
?>
我认为你的想法是正确的。但是，您有一个轻微的语法问题。
我相信你的变量没有计算为字符串，MySQL 期待字符串。请尝试以下操作：
$query = "UPDATE instructor SET
instructor = '{$fullname}',
gender = '{$gender}',
dob = '{$dob}',
nationality = '{$nationality}',
mobile = {$mobile},
emailadd = '{$emailadd}',
address = '{$address}',
zipcode = {$zipcode},
username = '{$username}',
WHERE instructorid= {$idupdate}";
请注意，单引号 （'） 位于大括号 （{}） 之外。您必须在MySQL需要字符串的任何位置添加引号（列类型为text/varchar/etc..）。
-更新-
正如sverri所说，它们实际上被视为字符串。这里的问题是 php 的插值有微小的变化：您可以使用 {$variable} 或简单地在双引号 （"） 字符串中$variable。在你的帖子中，你有{'$variable'}使用$variable方法进行插值，并且只是将大括号视为字符串的一部分，当你打算用 PHP 替换它们时。