Question

所以我有一个方法，只要点击一个按钮就会关闭。按钮是根据数据库中表的行数生成的。它们中的每一个都有一个唯一的id，它与数据库中的特定项id设置相同（这将使签入和退出项更容易。）所以方法看起来像这样

function calculate(event) {
  console.log(event.target.id);
  var id = event.target.id;
  var valueofbut = event.target.value;
  if(valueofbut==="Check Out"){
    document.getElementById(id).value = "Check In";
 $.ajax({
  type: "POST",
  url: "checkitemin.php",
  }).done(function( msg ) {
  alert( "Item has been successfully checked out" + msg );
  }); 

  }else{
    document.getElementById(id).value = "Check Out";
     $.ajax({
  type: "POST",
  url: "checkitemout.php",
  }).done(function( msg ) {
  alert( "Item has been successfully checked in" + msg );
  }); 

  }
}

我对php和ajax相当新，如何将变量id传递给ajax以便在checkitemout.php中使用。

我的checkitemout.php看起来像这样

<?php//basically is there any way to use that variable id from the method in this document..?
 session_start();
 require_once 'Dbconnect.php';
 $res=mysql_query("SELECT * FROM members WHERE memberid=".$_SESSION['user']);
 $userRow=mysql_fetch_array($res);
 $usersemail = $userRow['email'];
 $sql = "UPDATE inventory SET name ='$usersemail' WHERE ";
 mysql_query($sql);

?>

任何帮助都会很棒，谢谢大家。

Answer 1

Javascript

function calculate(event) {
  console.log(event.target.id);
  var id = event.target.id;
  var valueofbut = event.target.value;
  if(valueofbut==="Check Out"){
    document.getElementById(id).value = "Check In";
 $.ajax({
  type: "POST",
  url: "checkitemin.php",
  data: {idparam:id}, //send value of variable id
  }).done(function( msg ) {
  alert( "Item has been successfully checked out" + msg );
  }); 

  }else{
    document.getElementById(id).value = "Check Out";
     $.ajax({
  type: "POST",
  url: "checkitemout.php",
  data: {idparam:id},
  }).done(function( msg ) {
  alert( "Item has been successfully checked in" + msg );
  }); 

  }
}

PHP

$id=$_POST['idparam']

尝试将项目ID用于ajax

1 个答案: