PHP AJAX多表单不插入到数据库
PHP AJAX多表单不插入到数据库
提问于 2018-06-21 04:22:39
代码运行正常，但当我单击submit时，它不会插入到数据库中，也不会返回任何sql错误。这是我的索引代码：
<?php
require_once("functions/functions.php");
$db_handle = new DBController();
$query ="SELECT * FROM country";
$results = $db_handle->runQuery($query);
error_reporting();
?>
<html>
<head>
<title>Kickoff360 - Add Match Fixtures</title>
<link rel="stylesheet" href="/style/bootstrap.min.css" >
<link rel="stylesheet" href="/style/style.css" >
<script src="/style/bootstrap.min.js"></script>
<script src="/style/jquery.min.js"></script>
</head>
<body>
<div class="container"><h3>Add Match Fixtures</h3>
<div class="omenu">P/S: Use the <b>Add More</b> button to add more input fields, <b>X</b> to remove input fields</div>
<br />
<div class="form-group">
<form name="add_team" class="form-inline" id="add_team">
<div class="table-responsive">
<table class="table table-bordered" width="100%" id="fieldset">
<tr class="danger">
<td width="10%" class="trH"><span>LEAGUE</span></td>
<td width="60%" class="trH"><span>HOME & AWAY TEAMS</span></td>
<td width="3%" class="trH"><span>DATE</span></td>
<td width="3%" class="trH"><span>TIME</span></td> </tr></thead>
<tr id="row1"><td width="10%">
<select name="league[]" class="form-control" onChange="getState(this.value);">
<option value="">Select League</option>
<?php foreach($results as $country) { ?><option id="<?php echo $country["id"]; ?>" value="<?php echo $country["id"]; ?>"><?php echo $country["country_name"]; ?></option>
<?php } ?></select></td><td width="60%">
<select class="form-control" name="home[]" id="home1" class="select"><option value="">HOME TEAM</option></select>
<span class="btn btn-danger">VS</span>
<select class="form-control" name="away[]" id="away1" class="select"><option value="">AWAY TEAM</option></select>
</td><td width="3%"><input type="date" name="date[]" class="form-control"></td> <td width="3%"><input type="time" name="time[]" class="form-control"></td>
<td width="20%"><button type="button" name="add" id="add" class="btn btn-success" style="margin-left: 1px; margin-right: -1;">Add</button><span></span><button type="button" class="btn btn-success" style="display:inline-block;" id="submit" name="submit">Submit</button></td></tr>
</table>
</div>
</form>
</div>
</div>
</body>
</html>
<script>
var i=1;
function getState(val) {
var home = $("#home"+i+"");
var away = $("#away"+i+"");
$.ajax({
type: "POST",
url: "name.php",
data: 'country_id='+val,
success: function(data){
$(home).html(data).serialize();
$(away).html(data).serialize();
}
});
}
function selectCountry(val) {
$("#search-box").val(val);
$("#suggesstion-box").hide();
}
$(document).ready(function (){
$('#add').click(function(){
i++;
var home = $("#home"+i+"");
var away = $("#away"+i+"");
$(fieldset).append('<tr id="row'+i+'"><td width="10%"><select name="country" class="form-control" onChange="getState(this.value);"><option value="">Select Country</option><?php foreach($results as $country) { ?><option value="<?php echo $country["id"]; ?>"><?php echo $country["country_name"]; ?></option><?php } ?></select></td><td width="50%"><select name="home[]" id="home'+i+'" class="form-control"><option value="">HOME TEAM</option></select> <span class="btn btn-danger">VS</span> <select name="away[]" id="away'+i+'" class="form-control"><option value="">AWAY TEAM</option></select></td><td width="3%"><input type="date" class="form-control"></td> <td width="3%"><input type="time" class="form-control"></td><td><button type="button" name="remove" id="'+i+'" class="btn btn-danger btn_remove">X</button></td></tr>');
});
$(document).on('click', '.btn_remove', function(){
var button_id = $(this).attr("id");
$('#row'+button_id+'').remove();
});
$('#submit').click(function(){
$.ajax({
url:"name.php",
method:"POST",
data:$('#add_team').serialize(),
success:function(data)
{
alert(data);
//$('#add_team')[0].reset();
}
});
});
});
</script>
这是我的name.php
<?php
require_once('C:\wamp645\www\fixtures\functions\functions.php');
$db_handle = new DBController();
if(!empty($_POST["country_id"])) {
$query = "SELECT * FROM states WHERE countryID = '" . $_POST["country_id"] . "'";
$results = $db_handle->runQuery($query);
?>
<option value="">SELECT TEAM</option>
<?php
foreach($results as $state) {
?>
<option value="<?php echo $state["name"]; ?>"><?php echo $state["name"]; ?></option>
<?php
}
}
$connect = mysqli_connect("localhost", "root", "", "k_fixtures");
$number = count($_POST["home"]);
if($number >= 0)
{
for($i=0; $i<$number; $i++)
{
$league = $_POST['league'][$i];
$home = escapeStr($_POST["home"][$i]);
$away = escapeStr($_POST["away"][$i]);
$date = escapeStr($_POST['date'][$i]);
$time = escapeStr($_POST['time'][$i]);
$error = array();
if(empty($league)) {
$error = "This field cannot be blank";
}
if(empty($home)) {
$error = "This field cannot be blank";
}
if(empty($away)) {
$error = "This field cannot be blank";
}
if(empty($date)) {
$error = "This field cannot be blank";
}
if(empty($time)) {
$error = "This field cannot be blank";
}
$sql = "INSERT INTO k_fixtures SET
league = '$league',
team1 = '$home',
team2 = '$away',
dates = '$date',
times = '$time'
";
$sql2 = querySql($sql); or die(mysqli_error($connect));
}
}
if($sql2) {
echo "Data Inserted";
}
else
{
echo "Data not Inserted";
}
注意:在我使用文本输入时，函数querySql()已经在functions.php中声明了代码存储到我的数据库中，直到我最近添加了select选项。
回答 2
Stack Overflow用户
发布于 2018-06-21 04:42:39
$sql = "INSERT INTO k_fixtures (league,team1,team2,dates,times)
VALUES ('$league','$home','$away','$date','$time')";我想这就是你要找的嵌入物
Stack Overflow用户
发布于 2018-06-21 07:49:04
insert语句是正确的，但将insert放入如下条件中:-
if($error == ''){
$sql = "INSERT INTO k_fixtures SET
league = '$league',
team1 = '$home',
team2 = '$away',
dates = '$date',
times = '$time'";
$sql2 = querySql($sql); or die(mysqli_error($connect));
if($sql2) {
echo "Data Inserted";
}
else{
echo "Data not Inserted";
}
}页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/50956464
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