Question

我在使用php插入mysql数据库时遇到问题。特别是这段代码给了我一些问题：

$sql = "INSERT INTO '$mysql_database'.'$UsersTable' ('firstName', 'lastName', 'password', 'email','userType')VALUES ('$firstName', '$lastName', '$password','$email','$userType')";

我似乎无法在此处找到语法错误。给我的所有信息似乎都是正确的。我哪里错了？

Answer 1

试试这个

$sql = "INSERT INTO $mysql_database.$UsersTable (firstName, lastName, password, email,userType)VALUES ('$firstName', '$lastName', '$password','$email','$userType')";

从表格列中删除引用

Answer 2

尝试此查询

$sql = "INSERT INTO $mysql_database.$UsersTable (firstName, lastName, password, email,userType)VALUES ('$firstName', '$lastName', '$password','$email','$userType')";

Answer 3

您好，您无法使用表格列名中的单个引用，因此请将其删除

<?php
    session_start();
    $user_id = $_SESSION['uid']; 
    include('db.php');

    $sql = "SELECT * from registered_users";
    $query = mysql_query($sql) or die(mysql_error());

    $result = mysql_query("SELECT * FROM user_files_public where uid = $user_id");

?>      

<!DOCTYPE html>
<html ng-app>
    <head>
        <meta http-equiv="Content-Type" content="text/html; charset=UTF-8">
        <meta name="viewport" content="width=device-width, initial-scale=1" />
        <link rel="stylesheet" type="text/css" href="style.css">
        <script src="http://code.jquery.com/jquery-2.2.0.min.js"></script>

<script src = "http://code.jquery.com/jquery-1.9.1.js"></script>
<script type="text/javascript" src="ajax-jaquery.js"></script>
<script type="text/javascript">

$(function(){
  $(".file").click(function(){
    $(".userlist").click(function(){
      var elementUser = $(this);
      var elementFile = $(".file").$(this);

      var name = elementUser.attr("id");
      var id = elementFile.attr("name");

      var info='name='+name+'&id='+id;

      $.ajax({
        type: "POST",
        url: "share.php",
        data: info,
        success: function(){
       }
     });
      $(this).parents(".show").animate({ backgroundColor: "#003" }, "slow")
  .animate({ opacity: "hide" }, "slow");
return true;
    });
  });
});

</script>

<style>

        .ontop_share {
        z-index: 999;
        width: 900px;
        height: 900px;
        top: 0;
        left: 0;
        display: none;
        position: absolute;       
        background-color: #cccccc;
        color: #aaaaaa;
        opacity: 1.9;
        filter: alpha(opacity = 50);
      }
      #popup_share {
        width: 900px;
        height: 900px;
        position: absolute;
        color: #000000;
        border: 2px solid red;
        background-color: #ffffff;
        top: 50%;
        left: 50%;
        margin-top: -100px;
        margin-left: -150px;
      }

 </style>
<script type="text/javascript">
      function pop(div) {
        document.getElementById(div).style.display = 'block';
      }
      function hide(div) {
        document.getElementById(div).style.display = 'none';
      }

</script>

   </head>

    <body>

<div id="popDiv_share" class="ontop_share">

<div id="popup_share">
      <?php

  while($row = mysql_fetch_array($query)){
    $id = $row['id'];
    $name =$row['first_name'];?>

     <a href="" class="userlist" id="<?php echo $id;?>"><?php echo $id.' '.$name;?></a><?php  
    echo '<br/>';


  }
  ?>
  </div>  
            <a href="#" onClick="hide('popDiv_share')">Close</a>

</div>

    <table id="repos" align="center" cellspacing="20px">


            <thead>

                <tr>
                    <th>File name</th>
                    <th>Share</th>
                </tr>
            </thead>


                   <?php

                while($row = mysql_fetch_array($result)) { 
                    $id=$row['id'];
                    $user_id = $row['uid'];
                    $filename =$row['user_file_name'];


           ?> 

    <tbody>
        <tr>
            <td id ="actions"><?php echo $filename; ?></td>

            <td id ="actions">
                <a name="<?php echo $filename;?>" class="file" href="#" onClick="pop('popDiv_share')">Share</a>
            </td>


        </tr>
    </tbody>

    <?php     
          }                                
    ?> 
    </table>

</body>
</html>

否则使用这种方式

$sql = "INSERT INTO $mysql_database.$UsersTable (firstName, lastName, password, email,userType)VALUES ('$firstName', '$lastName', '$password','$email','$userType')";

php插入sql的问题

3 个答案: