使用PHP错误列出和删除MySQL中的记录
[英]List & Delete records in mysql using php error
问题描述
我已经写了一个PHP代码。 我无法从数据库中检索结果，也无法删除。 在提交时，它只给出一个空白页而不会引发任何错误。 我对此并不陌生，所以即使您认为这是一个愚蠢的问题，也请回复。 请参考代码，并向我提出一些更改，这些更改将使我的代码正常工作。
<!DOCTYPE html> <html> <head> <link rel="stylesheet" type="text/css" href="style.css"> </head> <body> <title>EDIT SCREEN</title> <form action="test4.php" method="post"> <ul> <li> Employee ID:</br> <input type="text" name="eid"> </li> <li> <input type="submit" value="SUBMIT"> </li> </ul> </form> </body> </html> //test.php <?php define('DB_NAME', 'test'); define('DB_USER', '**'); define("DB_PASSWORD", '**'); define('DB_HOST', 'localhost'); $link = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD); $db_selected = mysql_select_db(DB_NAME, $link); if(isset($_POST['ok'])){ $value1 = $_POST['eid']; $res = mysql_query("SELECT * from 'add' WHERE empid = '".$value1."'"); echo "<table border='1'> <tr><th>Name</th> <th>EmployeeID</th><th>Address</th></tr>"; while($row = mysql_fetch_array($res)) { echo "<tr>"; echo "<td>" . $row['name'] . "</td>"; echo "<td>" . $row['empid'] . "</td>"; echo "<td>" . $row['desig'] . "</td>"; echo "<td><a href='test5.php?del=$row[empid]'>Delete</a></td>"; echo "</tr>"; } echo "</table>"; } if (!mysql_query($sql)) { die('Error: ' . mysql_error()); } mysql_close(); ?> //test5.php <?php define('DB_NAME', 'test'); define('DB_USER', ''); define("DB_PASSWORD", ''); define('DB_HOST', 'localhost'); $link = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD); $db_selected = mysql_select_db(DB_NAME, $link); $value1 = $_POST['del']; mysql_query("DELETE FROM add WHERE empid = '$value1'") ?>
3 个解决方案
解决方案1
3 已采纳 2014-12-23 05:52:21
<?php
if(isset($_POST['ok'])){
$value1 = $_POST['eid'];
$res = mysql_query("SELECT * from `add` WHERE empid = '".$value1."'");
echo "<table border='1'>
<tr><th>Name</th>
<th>EmployeeID</th><th>Address</th></tr>";
while($row = mysql_fetch_array($res))
{
echo "<tr>";
echo "<td>" . $row['name'] . "</td>";
echo "<td>" . $row['empid'] . "</td>";
echo "<td>" . $row['add'] . "</td>";
echo "</tr>";
}
echo "</table>";
}
?>
New Code
<!DOCTYPE html>
<html>
<head>
<link rel="stylesheet" type="text/css" href="style.css">
</head>
<body>
<title>EDIT SCREEN</title>
<form action="result.php" method="post">
<ul>
<li>
Employee ID:</br>
<input type="text" name="eid">
</li>
<li>
<input type="submit" value="SUBMIT" name="ok">
</li>
</ul>
</form>
</body>
</html>
<?php
mysql_connect("localhost","root","");
mysql_select_db("fdd");
if(isset($_POST['ok'])){
$value1 = $_POST['eid'];
$res = mysql_query("SELECT * from `jobs` WHERE id = '".$value1."'");
echo "<table border='1'>
<tr><th>Name</th>
<th>EmployeeID</th><th>Address</th></tr>";
while($row = mysql_fetch_array($res))
{
echo "<tr>";
echo "<td>" . $row['job_date'] . "</td>";
echo "<td>" . $row['client_code'] . "</td>";
echo "<td>" . $row['department'] . "</td>";
echo "</tr>";
}
echo "</table>";
}
?>
output
解决方案2
0 2014-12-23 05:48:43
在查询数据库之前，您必须先建立连接。 按照下面的代码片段示例，首先创建与mysql的连接，然后查询数据库。
<?php
$con=mysqli_connect("localhost","my_user","my_password","my_db");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
// Perform queries
mysqli_query($con,"SELECT * FROM Persons");
mysqli_query($con,"INSERT INTO Persons (FirstName,LastName,Age)
VALUES ('Glenn','Quagmire',33)");
mysqli_close($con);
?>
解决方案3
0 2014-12-23 06:10:37
试试这个代码，不要直接在mysql查询中使用id，直接在mysql查询中使用id之前，请从mysql_real_escape_string传递它，它将从sql注入中清除id。 请参阅SQL注入REF 。
<!DOCTYPE html>
<html>
<head>
<link rel="stylesheet" type="text/css" href="style.css">
</head>
<body>
<title>EDIT SCREEN</title>
<form action="test4.php" method="post">
<ul>
<li>Employee ID:</br><input type="text" name="eid"></li>
<li><input type="submit" value="SUBMIT"></li>
</ul>
</form>
</body>
</html>
********* test4.php file ***********
<?php
$value1 = mysql_real_escape_string($_POST['eid']);
if($value1)
{
$Connection = mysql_connect('localhost','root','') or die(mysql_error());
$ConnectionDB = mysql_select_db($Connection) or die(mysql_error());
$res = mysql_query("SELECT * from `add` WHERE empid = '".$value1."'") or die(mysql_error());
echo "<table border='1'>
<tr>
<th>Name</th>
<th>EmployeeID</th>
<th>Address</th>
</tr>";
while($row = mysql_fetch_array($res))
{
echo "<tr>";
echo "<td>" . $row['name'] . "</td>";
echo "<td>" . $row['empid'] . "</td>";
echo "<td>" . $row['add'] . "</td>";
echo "</tr>";
}
echo "</table>";
}
注意：当您的代码工作时，请使用“;”删除此“或die（mysql_error（）） ”。 它仅在生产模式下使用。
列出使用php的mysql中的记录，但是总共只出现一条记录？
[英]List records in mysql using php but only one record appear in all?
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