0
我正在建立一个朋友请求系统，但我的问题是如何获取每个用户的值，这是我的代码中的id，但是到目前为止，并且只返回第一个用户ID，即使你尝试添加其他朋友时，也只能得到我的数据库中的第一个用户ID，请有人应该修复我的代码。如何获取表单提交的每个输入值
users.php
<div class="users_b">
<?php
include 'db.php';
$sq = "select * from alert_users_account";
$query = mysqli_query($con,$sq);
while($row = mysqli_fetch_assoc($query)){
?>
<div class="user_dis_p">
<div id="user_img"><a href="alert_profile.php?id=<?php echo $row['id']?>"><img src="alert<?php echo $row['photo']; ?>">
</a></div> <div id="user_fs">
<a href="alert_profile.php?id=<?php echo $row['id']?>">
<?php echo $row['firstname']." "." "." "." ".$row['surname'];?></a>
<form id="f_form">
<input type="text" name="friend_id" id="friend_id" value="<?php echo $row['id']?>">
<input type="submit" id="add_user" value="ADD" onclick="return request()">
</form>
</div><?php }?>
</div>
</div>
<div id="msg"></div>
<script>
function request(){
var frnd = document.getElementById("friend_id").value;
alert(frnd);
$.ajax({
type:'get',
url:'user_request.php',
data:{
frnd:frnd
},
cache:false,
success: function(message){
$("#msg").html(message);
}
});
return false;
}
</script>
user_request.php
<?php
session_start();
include 'db.php';
if(isset($_SESSION['email'])){
$eml = $_SESSION['email'];
$sq = $con->prepare("select id from alert_users_account where email_phone=?");
$sq->bind_param('s',$eml);
$sq->execute();
$res = $sq->get_result();
$ro = $res->fetch_assoc();
$user = $ro['id'];
if($_GET['frnd']){
echo $id = $_GET['frnd'];
$sql = $con->prepare("select * from alert_users_account where id=?");
$sql->bind_param('i',$id);
$sql->execute();
$result = $sql->get_result();
$row = $result->fetch_assoc();
$frnd = $row['id'];
$sql = $con->prepare("insert into friend (user_id,friend_id)values (?,?)");
$sql->bind_param('ss',$user,$frnd);
$sql->execute();
$sql->close();
echo "successfully inserted!";
}else{
echo 'error';
}
}else{
echo 'you cant add a friend!';
}
?>
我也尝试使用POST方法，但二者之间仍然给出相同的反馈