我正在为即将举行的活动创建一个邀请卡应用程序。我的代码成功地将数据插入到具有表名数据的名为booking的mysql数据库中。但是检索存在问题。当我填写表单并提交时,它会将数据保存在db中但不生成任何内容。它给出了以下错误:
致命错误:在第44行的C:\ xampp \ htdocs \ booking \ index.php中调用资源上的成员函数query()
这是我的代码,请告诉我如何解决此问题。我将非常感谢你。
<html>
<body>
<?php
if(isset($_POST['add'])){
$dbhost = 'localhost';
$dbuser = 'root';
$dbpass = '';
$conn = mysql_connect($dbhost, $dbuser, $dbpass);
if(! $conn){
die('Could not connect: ' . mysql_error());
}
if(! get_magic_quotes_gpc()){
$emp_name = addslashes($_POST['emp_name']);
$emp_fname = addslashes($_POST['emp_fname']);
$emp_cnic = addslashes($_POST['emp_cnic']);
$emp_address = addslashes($_POST['emp_address']);
} else {
$emp_name = addslashes($_POST['emp_name']);
$emp_fname = addslashes($_POST['emp_fname']);
$emp_cnic = addslashes($_POST['emp_cnic']);
$emp_address = addslashes($_POST['emp_address']);
}
$sql = "INSERT INTO data ". "(CNIC, Name, FatherName, PostalAddress) " .
"VALUES('$emp_cnic', '$emp_name', '$emp_fname', '$emp_address')";
mysql_select_db('booking');
$retval = mysql_query($sql, $conn);
if(! $retval) {die('Could not enter data: ' . mysql_error());}
?>
<table border=2>
~~~~~~Your Invitation Card~~~~~
<tr><td>Your Name</td><td><?php
$sql = "SELECT name FROM data";
$result = $conn->query($sql);
echo $result;
?></td></tr><br>
<tr><td>Your Father Name</td><td>
$sql = "SELECT fname FROM data";
$result = $conn->query($sql);
echo $result;
?></td></tr><br>
<tr><td>Your CNIC Number</td><td>
$sql = "SELECT cnic FROM data";
$result = $conn->query($sql);
echo $result;
?></td></tr><br>
<tr><td>Your Postal Address</td><td>
$sql = "SELECT address FROM data";
$result = $conn->query($sql);
echo $result;
?></td></tr><br>
<tr><td>You are informed to approach Location XA-55 at 1800 Thursday with print of this
Invitation card to paticipate in the function. </td></tr><br>
</table>
<?php
mysql_close($conn);
} else {
?>
<form method = "post" action = "<?php $_PHP_SELF ?>">
Name: <input type="text" name="emp_name" id="emp_name"><br>
Father Name: <input type="text" name="emp_fname" id="emp_fname"><br>
CNIC: <input type="text" name="emp_cnic" id="emp_cnic"><br>
Address: <input type="text" name="emp_address" id="emp_address"><br>
<input type="submit" name="add" id="add" value="Submit">
<?php
}
?>
</body></html>
答案 0 :(得分:1)
更改
<!DOCTYPE html>
<html lang="en">
<head>
<link rel="stylesheet" type="text/css" href="style.css">
<link rel="stylesheet" type="text/css" href="css/bootstrap.css">
<script src="drag.js"></script>
</head>
<body>
<div class="container">
<div class="row">
<div class="col-md-3" id="1box">
<img id="img1" src="images/image1.jpg">
</div>
<div class="col-md-3" id="2box">
<img id="img2" src="images/image2.jpg">
</div>
<div class="col-md-3" id="3box">
<img id="img3" src="images/image4.jpg">
</div>
<div class="col-md-3" id="4box">
<img id="img4" src="images/image5.jpg">
</div>
</div>
</div>
<div class="row">
<center><div class="col-md-push-4 col-md-4 mainbox" ondrop="drop(event)" id="mainbox">
<img id="centerimg" src="images/center.png">
</div></center>
</div>
</body>
</html>
要
$result = $conn->query($sql);
了解更多信息click here
答案 1 :(得分:0)
我认为您应该使用mysql_query而不是$ conn-&gt; query
答案 2 :(得分:0)
我很瘦我在你的代码中发现了两个错误。
你应该使用
mysql_query($sql,$conn);
而不是(之前提到的)
$result = $conn->query($sql);
你在html表格中错过了几个开放的php标签。
尝试以下代码,如果有效,请告诉我。
if(! $conn){
die('Could not connect: ' . mysql_error());
}
if(! get_magic_quotes_gpc()){
$emp_name = addslashes($_POST['emp_name']);
$emp_fname = addslashes($_POST['emp_fname']);
$emp_cnic = addslashes($_POST['emp_cnic']);
$emp_address = addslashes($_POST['emp_address']);
} else {
$emp_name = addslashes($_POST['emp_name']);
$emp_fname = addslashes($_POST['emp_fname']);
$emp_cnic = addslashes($_POST['emp_cnic']);
$emp_address = addslashes($_POST['emp_address']);
}
$sql = "INSERT INTO data ". "(CNIC, Name, FatherName, PostalAddress) " .
"VALUES('$emp_cnic', '$emp_name', '$emp_fname', '$emp_address')";
mysql_select_db('booking');
$retval = mysql_query($sql, $conn);
if(! $retval) {die('Could not enter data: ' . mysql_error());}
?>
<table border=2>
~~~~~~Your Invitation Card~~~~~
<tr><td>Your Name</td><td><?php
$sql = "SELECT name FROM data";
$result = mysql_query($sql,$conn);
echo $result;
?></td></tr><br>
<tr><td>Your Father Name</td><td>
<?php
$sql = "SELECT fname FROM data";
$result = mysql_query($sql,$conn);
echo $result;
?></td></tr><br>
<tr><td>Your CNIC Number</td><td>
<?php
$sql = "SELECT cnic FROM data";
$result = mysql_query($sql,$conn);
echo $result;
?></td></tr><br>
<tr><td>Your Postal Address</td><td>
<?php
$sql = "SELECT address FROM data";
$result = mysql_query($sql,$conn);
echo $result;
?></td></tr><br>
<tr><td>You are informed to approach Location XA-55 at 1800 Thursday with print of this
Invitation card to paticipate in the function. </td></tr><br>
</table>
<?php
mysql_close($conn);
} else {
?>
<form method = "post" action = "<?php $_PHP_SELF ?>">
Name: <input type="text" name="emp_name" id="emp_name"><br>
Father Name: <input type="text" name="emp_fname" id="emp_fname"><br>
CNIC: <input type="text" name="emp_cnic" id="emp_cnic"><br>
Address: <input type="text" name="emp_address" id="emp_address"><br>
<input type="submit" name="add" id="add" value="Submit">
<?php
}
?>
</body></html>