任何人都可以帮助我...如何在ajax中放置条件警告对话框,如果查询中的数据已成功保存或数据已保存。
我想要做的是,如果保存了查询,如果数据已经保存,则会弹出一个警告框。
脚本代码:
<script type="text/javascript">
$(document).ready(function () {
$('#updates').click(function (e) {
e.preventDefault();
var data = {};
data.region_text = $('#t_region').val();
data.town_text = $('#t_town').val();
data.uniq_id_text = $('#t_uniq_id').val();
data.position_text = $('#t_position').val();
data.salary_grade_text = $('#t_salary_grade').val();
data.salary_text = $('#t_salary').val();
for(var $x=1;$x<=15;$x++) {
data['id'+$x+'_text'] = $('#id'+$x).val();
data['aic'+$x+'_text'] = $('#aic'+$x).val();
data['name'+$x+'_text'] = $('#name'+$x).val();
data['optA'+$x+'_text'] = $('#optA'+$x).val();
data['optB'+$x+'_text'] = $('#optB'+$x).val();
data['optC'+$x+'_text'] = $('#optC'+$x).val();
data['optD'+$x+'_text'] = $('#optD'+$x).val();
data['other_qual'+$x+'_text'] = $('#other_qual'+$x).val();
data['interview'+$x+'_text'] = $('#interview'+$x).val();
data['total'+$x+'_text'] = $('#total'+$x).val();
}
$.ajax({
type: "POST",
url: "insert.php",
data: data,
cache: false,
success: function (response) {
// We are using response to distinguish our outer data variable here from the response
}
});
});
});
</script>
insert.php代码:
<?php
include('../connection.php');
date_default_timezone_set('Asia/Manila');
$region = @$_POST['region_text'];
$town = @$_POST['town_text'];
$uniq_id = @$_POST['uniq_id_text'];
$position = @$_POST['position_text'];
$salary_grade = @$_POST['salary_grade_text'];
$salary = @$_POST['salary_text'];
$dupesql = "SELECT * FROM afnup_worksheet WHERE funiq_id = '$uniq_id'";
$duperow = mysql_query($dupesql);
if(mysql_num_rows($duperow) > 0){
exit;
}else{
for($n=1;$n<=15;$n++) {
$id = @$_POST['id'.$n.'_text'];
$aic = @$_POST['aic'.$n.'_text'];
$name = @$_POST['name'.$n.'_text'];
$optA = @$_POST['optA'.$n.'_text'];
$optB = @$_POST['optB'.$n.'_text'];
$optC = @$_POST['optC'.$n.'_text'];
$optD = @$_POST['optD'.$n.'_text'];
$other_qual = @$_POST['other_qual'.$n.'_text'];
$interview = @$_POST['interview'.$n.'_text'];
$total = @$_POST['total'.$n.'_text'];
if(!empty($name)){
$query = "INSERT INTO afnup_worksheet (faic,fregion,ftown,funiq_id,fposition,fsalary_grade,fsalary,fnl_name,edu_attain,experience,seminars,eligibility,other_qual,interview,ftotal,dateinputed)
VALUES
('$aic','$region','$town','$uniq_id','$position','$salary_grade','$salary','$name','$optA','$optB','$optC','$optD','$other_qual','$interview','$total',CURRENT_TIMESTAMP)";
$resource = mysql_query($query) or die(mysql_error());
}
}
}
?>
答案 0 :(得分:2)
只需从PHP返回该状态:
if(mysql_num_rows($duperow) > 0){
echo "1"; // Dup status
exit;
}else{
// All your else code.. echo must be the last thing inside your else block
echo "2"; // Saved status
}
然后在你的ajax success
回调中检查它:
$.ajax({
type: "POST",
url: "insert.php",
data: data,
cache: false,
success: function (response) {
if (Number(response) == 1)
{
alert("Dup message");
}
else
{
alert("Saved message");
}
}
});
答案 1 :(得分:1)
而不是退出;在你的骗局中,你可以回应“重复”。你也应该在$ resource之后删除die()并添加if($ resource)echo“ok”;否则回应“错误”;
然后在你的javascript成功函数(响应)中你可以做if(response ==“...”)echo duplicate;否则如果......
这只是基本的解释,但它足以指出你正确的方向。