Php ajax回调方法,但返回不成功
问题:我可以将值传递给ajax方法并成功实现selectsql(在网络中可以看到结果),但它不会返回任何结果(我的结果是alert(“successful”)。这有什么问题 HTML: PHP:Php ajax回调方法,但返回不成功,php,html,ajax,callback,Php,Html,Ajax,Callback,问题:我可以将值传递给ajax方法并成功实现selectsql(在网络中可以看到结果),但它不会返回任何结果(我的结果是alert(“successful”)。这有什么问题 HTML: PHP: $serverName = "localhost"; $username = "root"; $password = "root"; $dbName = "noticeboard"; $tbName1 = "notice"; $conn = new mysqli($serv
$serverName = "localhost";
$username = "root";
$password = "root";
$dbName = "noticeboard";
$tbName1 = "notice";
$conn = new mysqli($serverName, $username, $password, $dbName);
checkStaffLogin();
function checkStaffLogin()
{
global $conn, $tbName1;
if ($conn->connect_errno)
{
echo "Fail to connect MYSQL database : " . $conn->connect_errno;
}
else {
$ID= $_POST["ID"];
$sql = "SELECT *
FROM $tbName1 where ID= '$ID'";
$result = $conn->query($sql);
if($result)
{
while($row =mysqli_fetch_assoc($result) )
{
$table_data[]= array("Name"=>$row['Name'],"Password"=>$row['Password']);
}
echo json_encode($table_data);
}
请尝试此查询
$table_data=[];
$sql = "SELECT * FROM ".$tbName1." where ID = ".$ID;
注意:使用preperd语句
并将此标题附加到php标题('Content-Type:application/json');
要循环结果,请使用
success: function (data) {
$.each(data,function(i,v){
console.log(v.Name)
});
}
注意:数据应如下所示:
data: {ID: ID},
在JS中尝试此操作通过删除这些“ID”检查语法中的数据。尝试以下操作
function selectupdate_Server(id)
{
var ID= id;
$.ajax({
url: "abc.php",
type: 'POST',
data: {ID: ID},
dataType: 'json',
success: function (data) {
alert ('Success');
}
});
}
您可以尝试这个Javascript函数
function selectupdate_Server(id)
{
var ID= id;
console.log(ID); --->Successfully logged
$.ajax({
type: 'POST',
data: {ID: ID},
url: "abc.php",
dataType: 'json',
success: function(resultss){
try{
alert ('Success');
}catch(e)
{
alert(JSON.stringify(e)+"Catch error");//any Error in Your php file display alert.where,what error
}
},
error: function(e){
alert(JSON.stringify(e)+"error");//any Error in Your php file display alert.where,what error
}
});
}
这个脚本在php文件中的错误会提醒您检查并更正行号和文件名
您可以这样编写Php代码
checkStaffLogin();
function checkStaffLogin()
{
$serverName = "localhost";
$username = "root";
$password = "root";
$dbName = "noticeboard";
$tbName1 = "notice";
$conn = mysqli_connect($serverName, $username, $password, $dbName);
if (mysqli_connect_errno($conn))
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
else {
$ID= $_POST["ID"];
$table_data=array();
$sql = "SELECT * FROM $tbName1 where ID= '$ID'";
$result = mysqli_query($conn,$sql) or die(mysql_error());
if($result)
{
while($row =mysqli_fetch_assoc($result) )
{
$table_data[]= array("Name"=>$row['Name'],"Password"=>$row['Password']);
}
echo json_encode($table_data);
}
}
}
您可以这样尝试add
exit;
在php函数的末尾,您的php代码可能包含错误。请检查浏览器控制台或向我们显示整个php代码未定义的索引:ID&未定义的变量:table_data但我在“console.log(ID)”之前有检查ID@soul299数据库中的ID是如何命名的?SW registed undefineddo您有一个名为SW的变量吗?我没有SW variablepost此ajax的所有php代码,表结构mysqli_fetch_assoc()希望参数1是mysqli_result,在&&&&Undefined variable中给出布尔值:table_data in“ID”之间有什么不同和ID,但在我使用“”之前,它太有效了{“readyState”:4,“responseText”:“\n警告:mysqli\u query()希望参数1是mysqli,字符串在第27行的C:\\xampp\\htdocs\\it\\Admin\u Notice\u SelectUpdate.php
\n
\n警告:mysqli\u fetch\u assoc()中给出预期参数1为mysqli_result,在第28行的C:\\xampp\\htdocs\\IT\\Admin_Notice_SelectUpdate.php“status”:200,“statusText”:“OK”}error显示数据库连接的完整php代码。我认为数据库连接中出现错误{“readyState”:4,“responseText”:“
\n警告:mysqli_error()第27行的C:\\xampp\\htdocs\\IT\\Admin\u Notice\u SelectUpdate.php中正好需要1个参数,0,在第27行
\n,“状态”:200,“状态文本”:“OK”}错误行27:$result=mysqli\u query($conn,$sql)或die(mysqli\u error());您有teamviewer吗?
function selectupdate_Server(id)
{
var ID= id;
console.log(ID); --->Successfully logged
$.ajax({
type: 'POST',
data: {ID: ID},
url: "abc.php",
dataType: 'json',
success: function(resultss){
try{
alert ('Success');
}catch(e)
{
alert(JSON.stringify(e)+"Catch error");//any Error in Your php file display alert.where,what error
}
},
error: function(e){
alert(JSON.stringify(e)+"error");//any Error in Your php file display alert.where,what error
}
});
}
checkStaffLogin();
function checkStaffLogin()
{
$serverName = "localhost";
$username = "root";
$password = "root";
$dbName = "noticeboard";
$tbName1 = "notice";
$conn = mysqli_connect($serverName, $username, $password, $dbName);
if (mysqli_connect_errno($conn))
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
else {
$ID= $_POST["ID"];
$table_data=array();
$sql = "SELECT * FROM $tbName1 where ID= '$ID'";
$result = mysqli_query($conn,$sql) or die(mysql_error());
if($result)
{
while($row =mysqli_fetch_assoc($result) )
{
$table_data[]= array("Name"=>$row['Name'],"Password"=>$row['Password']);
}
echo json_encode($table_data);
}
}
}