我有一个名为“添加”的按钮,它与列表中的每个人有关。
height: 200px
单击“添加”时,我想将人员详细信息添加到表中(工作正常),还要在我的数据库中的表中添加包含其名称和ID的新行(这是我的问题)。 / p>
我知道PHP不能在JS函数中工作。我也知道由于安全问题,我不应该从JS访问我网站上的数据库。
那么如何才能有 echo <td><input type=\"submit\" id=\"PlayerAdded".$id."\" value=\"Add\"
onclick=\"add('".$id."','".$name."','".$_SESSION['GameID']."');\"></input></td>";
调用添加到数据库的函数的按钮?
我尝试在PHP中编写我的函数,但是当我点击按钮时它没有做任何事情。请参阅下面的功能:
onclick
答案 0 :(得分:4)
删除变量之前的括号,这是为函数中的每个变量编写的。
在功能代码的末尾添加ajax
部分。
function add(id, name, game) {
var t = "</td><td>";
var str = "<tr id='Players" + id + "'><td>"
var ctr = "</td></tr>"
var PID = "<input type = 'hidden' name='ID" + id + "' value='" + id + "'></input>" + id;
var Pnam = "<input type='hidden' name='Name" + id + "' value='" + name + "'></input>";
var place = "<select name='Place" + id + "'><option value='17'>17th</option><option value='16'>16th</option><option value='15'>15th</option><option value='14'>14th</option><option value='13'>13th</option><option value='12'>12th</option><option value='11'>11th</option><option value='10'>10th</option><option value='9'>9th</option><option value='8'>8th</option><option value='7'>7th</option><option value='6'>6th</option><option value='5'>5th</option><option value='4'>4th</option><option value='3'>3rd</option><option value='2'>2nd</option><option value='1'>1st</option></select>";
var points = "<form action='leaderboards.php' method='post' target='_blank'><input type='hidden' name='playerid' value='" + id + "'></input><input type='hidden' name='name' value='" + name + "'></input><input type='submit' value='View'></form>";
var cash = "$<input name='Cash" + id + "' placeholder=' 0'></input>";
var ticket = "<select name='Ticket" + id + "'><option value='No'>No</option><option value='Yes'>Yes</option>";
var del = "<input type='button' value='Delete' onclick='remove(" + id + ")'> </input>";
$('#PlayerAdded').before(str + PID + t + Pnam + name + t + place + t + points + t + cash + t + ticket + t + del + ctr);
// making ajax call to insert.php and posting the data
$.ajax({
method: "POST",
url: "insert.php",
data: {
"id": id,
"name": name,
"game": game
},
beforeSend:function(){
// show something before data is saved to db.
}
}).done(function(msg) {
$("body").append(msg); //debugging purpose
}).fail(function( jqXHR, textStatus ) {
alert( "Request failed: " + textStatus );
});
}
<强> insert.php 强>
<?php
if(isset($_POST['id']) && isset($_POST['name']) && isset($_POST['game'])){
$id= $_POST['id'];
$name= $_POST['name'];
$game= $_POST['game'];
//write your connection logic
// never write bare queries there is a chance of sql injection.
//Instead use prepared statement, prepare your query then bind the above values to it, then excute.
//write your insert logic below.
}
?>
答案 1 :(得分:0)
您需要通过AJAX
发布要插入数据库的值,并将其附加到add()
函数:
$.post('insertnew.php', { id:id,name:name,game:game }, function(data){
alert(data);
});
<强> insertnew.php:强>
<?php
if(isset($_POST['id']) && isset($_POST['name']) && isset($_POST['game'])){
// sanitize your data here, then:
$id = $_POST['id'];
$name = $_POST['name'];
$game = $_POST['game'];
//connect to your db, or instead include your dbconnect.php
$link= new mysqli('localhost', 'my_user', 'my_password', 'world');
// check connection
if (mysqli_connect_errno()) {
printf("Connect failed: %s\n", mysqli_connect_error());
exit();
}
//use prepared statement, prepare your query then bind the above values to it, then excute
$stmt = mysqli_prepare($link, "INSERT INTO results VALUES (?, ?, ?)");
mysqli_stmt_bind_param($stmt, $id, $name, $game);
if(mysqli_stmt_execute($stmt)){
echo "Record has been added successfully";
}else{
echo "Sorry, record could not be inserted";
}
}
?>