Question

我试图从数据库中提取与文本框中的数据匹配的数据，而我正在使用的代码会引发错误。

我认为我的SQL查询有问题，但我不确定是什么。

错误是：

警告：mysql_fetch_array（）要求参数1为资源，第65行的C：\ wamp \ www \ search.php中给出布尔值

我的语法：

<html>
 <head>

   <link rel="stylesheet" type="text/css"  href="style.css"></link> 

   </head>
   <body>
     <form  action="" name="formdownload" method="post">

  <table>
  <tr><td colspan=2><h1>Domestic Flights</h1></td></tr></br>
      <td height=50> From:</td><td><input type="From" name="from" size=30/>
        <tr><td height=50>To: </td><td><input type="To" name="to" size=30/>
        <tr><td><input name="submit" type="submit" value ="Search"></tr></td>
        <table border="1" align="center" id="table1" cellpadding="0" cellspacing="0">
            <tr>
                <th>Flight No</th>
                <th>Flight Company</th>
                <th>Plane Type</th>
                <th>From</th>
                <th>To</th>
            </tr>
            <center>

            <?php

            $submit = @$_POST['submit'];
            $from = @$_POST['from'];
            $to = @$_POST['to'];
            if($submit)
            {
             $select=mysql_query("select * from flight where ffrom='$from'and To='$to'");
            while($row1=mysql_fetch_array($select))

             {
                $FlightNo = $row1['FlightNo'];
                $FlightCompany=$row1['FlightCompany'];
                $PlaneType = $row1['PlaneType'];
                $From =$row1['ffrom'];
                $To =$row1['To'];

            ?>


            <tr>

            <td width="90" align="center">
                <?php echo $FlightNo;?>
            </td>
            <td width="90" align="center">
                <?php echo $FlightCompany;?>
            </td>
            <td width="90" align="center">
                <?php echo $PlaneType;?>
            </td>
            <td width="90" align="center">
                <?php echo $From;?>
            </td>
            <td width="90" align="center">
                <?php echo $To;?>
            </td>

            </tr>
            <?php }}?>
            </table>

  </table>
 </form>
 </div>
  </div>
   </div>


  </div>

 </center>

</body>
 </html>

Answer 1

变量名称的行有错误：

$select=mysql_query("select * from logintbl where name='$uname'");

将其替换为：

$select=mysql_query("select * from logintbl where name='$name'");

编辑：因为你的字段被命名为“uname＆＃39;在您的数据库中，这是正确的代码（通过添加mysql_real_escape更安全一点）：

$select=mysql_query('select * from logintbl where uname="'.mysql_real_escape($name).'";');

REEDIT：您需要第二个参数，但在查询之前没有在PHP中检索它。你的代码应该是这样的：

<html>
<body>
    <form enctype="multipart/form-data" action="" name="formdownload" method="post">
        <table border="1" align="center" id="table1" cellpadding="0" cellspacing="0">
            <tr>
                <th>ID</th>
                <th>Name</th>
                <th>Password</th>
                <th>User Type</th>
            </tr>
            <center>
        <br/><br/><br/>
            ID: <input type="text" name="idno" /> <br/><br/>
            Name : <input type="text" name="name" /><br/><br/>
            Marks : <input type="text" name="marks" /><br/><br/>

            <input type="submit" name = "submit" value="submit">

        </center>

            <?php

            $submit = @$_POST['submit'];
            $name = @$_POST['name'];
            $idno = @$_POST['idno'];
            if($submit)
            {
             $select=mysql_query("select * from logintbl where uname='$name' and Id='$idno'");
            while($row1=mysql_fetch_array($select))

             {
                $id = $row1['ID'];
                $name=$row1['uname'];
                $pass = $row1['pass'];
                $type =$row1['type'];

            ?>


            <tr>

            <td width="300" align="center">
                <?php echo $id;?>
            </td>
            <td width="300" align="center">
                <?php echo $name;?>
            </td>
            <td width="300" align="center">
                <?php echo $pass;?>
            </td>
            <td width="300" align="center">
                <?php echo $type;?>
            </td>

            </tr>
            <?php }}?>
        </table>
    </form>
</body>

我仍强烈建议您在使用mysql_real_escape进行查询之前过滤变量。

从Web表单中的数据库读取时出错

1 个答案: