Question

我正在开发一个预订系统，我需要为每个预订生成一个唯一ID，并在用户点击＆＃34;添加新预订＆＃34;

时将其显示在模式内的表单上

我使用ajax和SQL来做到这一点，但有时我会得到重复的id。我所做的是每次点击按钮我都会将id插入表格

那是我的代码

 <script>
function showId() {

  if (window.XMLHttpRequest) {
    // code for IE7+, Firefox, Chrome, Opera, Safari
    xmlhttp=new XMLHttpRequest();
  } else { // code for IE6, IE5
    xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
  }
  xmlhttp.onreadystatechange=function() {
    if (this.readyState==4 && this.status==200) {
      document.getElementById("show-id").defaultValue=this.responseText;
    }
  }
  xmlhttp.open("GET","getid.php?=",true);
  xmlhttp.send();
}
</script>


<script>

$(function () {
    $('#openModal').on('click', function () {

  showId();



var mx_val=$("#show-idi").val();
        $.ajax({
            type: "POST",
            url: "insert-id.php",
            data: {mx_val:mx_val},
            dataType: "JSON",
            success: function(data) {
           /*  $("#message").html(data); */
             getDu();
          /*  $("p").addClass("alert alert-success");*/
            },
            error: function(err) {
  $("#message").html("Saved!");
            $("p").addClass("alert alert-success");

              console.log(err);
            }
        });






    });
});




</script>

插入-id.php

<?php


include('db.php');

$mx_val=$_POST['mx_val'];


require_once("dbcontroller.php");
$db_handle = new DBController();
$query ='SELECT max(mx_val) from mxvalue';
$results = $db_handle->runQuery($query);
foreach($results as $references) {
$maxvalue =  $references["max(mx_val)"] +1;

} 
$stmt = $DBcon->prepare("INSERT INTO mxvalue (mx_val) VALUES ('$maxvalue')");


if($stmt->execute())
{
  $res="Data Inserted Successfully:";
  echo json_encode($res);
}
else {
  $error="Not Inserted,Some Problem occured.";
  echo json_encode($error);
}

getid.php

<?php
$con = mysqli_connect('','','','');
if (!$con) { die('Could not connect: ' . mysqli_error($con));
} mysqli_select_db($con,"ajax_demo");
$sql="SELECT max(mx_val) from mxvalue";
$result = mysqli_query($con,$sql);
while($row = mysqli_fetch_array($result)) {
$maxid = $row["max(mx_val)"] +1;
?><?php echo $maxid; ?><?php } mysqli_close($con);
?>

Answer 1

你可以使用基于unix时间戳和一些随机数的id，或使用uuid生成器喜欢：

https://github.com/ramsey/uuid

通过这两种方式，你永远不会得到重复的ids

每次点击按钮时生成新的ID

1 个答案: