通过 PHP 从 MySQL 编辑数据
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Editing data from MySQL via PHP
提问by Lukas Pleva
I am running into a frustrating problem with a PHP script that's supposed to allow me to edit individual rows within my MySQL database.
我在使用 PHP 脚本时遇到了一个令人沮丧的问题，该脚本应该允许我编辑 MySQL 数据库中的各个行。
This is the file where all of the rows from the database are displayed; it works just like it's supposed to.
这是显示数据库中所有行的文件；它就像它应该的那样工作。
<table cellpadding="10">
<tr>
<td>ID</td>
<td>First Name</td>
<td>Last Name</td>
<td>E-mail</td>
<td>Phone</td>
</tr>
<?php
$username="username here";
$password="password here";
$database="database name here";
mysql_connect(localhost,$username,$password);
@mysql_select_db($database) or die( "Unable to select database");
$query="SELECT * FROM students";
$result=mysql_query($query);
mysql_close();
while ($row=mysql_fetch_array($result)){
echo ("<tr><td>$row[id]</td>");
echo ("<td>$row[first]</td>");
echo ("<td>$row[last]</td>");
echo ("<td>$row[email]</td>");
echo ("<td>$row[phone]</td>");
echo ("<td><a href=\"StudentEdit.php?id=$row[id]\">Edit</a></td></tr>");
}
echo "</table>";
?>
As you can see, each row has an "Edit" link that is supposed to allow the user to edit that individual student's data. Here, then, is StudentEdit.php:
如您所见，每一行都有一个“编辑”链接，应该允许用户编辑单个学生的数据。那么，这里是 StudentEdit.php：
<?php
$username="username";
$password="password";
$database="database";
mysql_connect(localhost,$username,$password);
$student_id = $_GET[id];
$query = "SELECT * FROM students WHERE id = '$student_id'";
$result = mysql_query($query);
$row = mysql_fetch_array($result);
mysql_close();
?>
<form method="post" action="EditStudentData.php" />
<table>
<tr>
<td><input type="hidden" name="id" value="<? echo "$row[id]" ?>"></td>
</tr>
<tr>
<td>First Name:</td>
<td><input type="text" name="first" value="<? echo "$row[first]" ?>"></td>
</tr>
<tr>
<td>Last Name:</td>
<td><input type="text" name="last" value="<? echo "$row[last]" ?>"></td>
</tr>
<tr>
<td>Phone Number:</td>
<td><input type="text" name="phone" value="<? echo "$row[phone]" ?>"></td>
</tr>
<tr>
<td>E-mail:</td>
<td><input type="text" name="email" value="<?echo "$row[email]" ?>"></td>
</tr>
</table>
</form>
When I execute this, however, I get the following error message:
但是，当我执行此操作时，我收到以下错误消息：
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home4/lukaspl1/public_html/StudentEdit.php on line 12
警告：mysql_fetch_array()：在第 12 行的 /home4/lukaspl1/public_html/StudentEdit.php 中提供的参数不是有效的 MySQL 结果资源
Any ideas what's wrong, and how to fix it?
任何想法出了什么问题，以及如何解决？
Thank you in advance!
先感谢您！
采纳答案by ryuusenshi
StudentEdit.php:you forgot to call @mysql_select_db($database) or die( "Unable to select database");before you executed the query
StudentEdit.php：@mysql_select_db($database) or die( "Unable to select database");在执行查询之前忘记调用
回答by satdev86
Remove the mysql_close from here
从这里删除 mysql_close
mysql_connect(localhost,$username,$password);
@mysql_select_db($database) or die( "Unable to select database");
$query="SELECT * FROM students";
$result=mysql_query($query);
mysql_close();
The code should mysql_connect(localhost,$username,$password);
代码应该是 mysql_connect(localhost,$username,$password);
@mysql_select_db($database) or die( "Unable to select database");
$query="SELECT * FROM students";
$result=mysql_query($query);
And moreover,you are going to use only key based resultset.. simply have mysql_fetch_assoc. And another suggestion would be instead of $row[id]..replace it with $row['id'].
而且，您将仅使用基于键的结果集.. 只需拥有 mysql_fetch_assoc。另一个建议是代替 $row[id].. 用 $row['id'] 替换它。
回答by ransilu.info Admin
This part of the code is wrong:
这部分代码有误：
$student_id = $_GET[id];
the correct code is
正确的代码是
$student_id = $_GET['id'];
code from www.ransilu.info
回答by ransilu.info Admin
this is the cod for edit the details dynamically
这是动态编辑细节的代码
<?php
include('db.php');
$id=$_REQUEST['id'];
$query="SELECT * FROM `camera details` WHERE id='".$id."'";
$result=mysqli_query($db,$query) or die(mysqli_error());
$row1=mysqli_fetch_assoc($result);
if(isset($_POST['submit'])&&(isset($_POST['new'])&&($_POST['new'])==1))
{
$id=$_REQUEST['id'];
foreach($_POST as $key=>$values)
{
if($key!="submit"){
$names[]=$key;
$val[]= "'".$values."'";
if($key!="new"){
$k[] = "`".$key."` = '".$values."'";
}
}
}
$output=implode(",",(array)($k));
//$v=implode(",",(array)($val));
// `name` = 'san'
$query="UPDATE `camera details` SET $output WHERE id='".$id."'";
$output=mysqli_query($db,$query) or die(mysqli_error($db));
if($output)
{
header('location:cameralist.php');
}
}
else{
?>
回答by ransilu.info Admin
this code was missing
此代码丢失
$select_db = mysql_select_db("$db_name");
if (!$select_db) {echo "Error Selecting Database";}
回答by Ashwini Agarwal
Try...
尝试...
echo ("<td><a href=\"StudentEdit.php?id=".$row['id']."\">Edit</a></td></tr>");
instead of
代替
echo ("<td><a href=\"StudentEdit.php?id=$row[id]\">Edit</a></td></tr>");
回答by indranatha madugalle
This code gives the option to add, search, edit and delete options. Thought it might to see all the options in one code.
此代码提供添加、搜索、编辑和删除选项的选项。认为可能会在一个代码中看到所有选项。
$searchedUsername = "";
$searchedEmail = "";
//registration (Add) function
if ( isset($_POST['stdregister'])){
$username = $_POST['stdusername'];
$password = $_POST['stdpassword'];
$email = $_POST['stdemail'];
$hashedPassword = md5($password);
$connection = mysqli_connect("localhost","root","","std");
$query = "INSERT INTO student VALUES ('$username','$hashedPassword','$email')";
if ( mysqli_query($connection,$query) == 1 ){
echo "Successfully saved";
}
else{
echo "<p style='color: #f00;'>There is an error</p>";
}
mysqli_close($connection);
}
//delete function
if ( isset($_POST['stddelete'])){
$username = $_POST['stddelusername'];
$connection = mysqli_connect("localhost","root","","std");
$query = "DELETE FROM student WHERE username LIKE '$username'";
mysqli_query($connection,$query);
echo mysqli_error($connection);
mysqli_close($connection);
}
//update function
if ( isset($_POST['stdupdate'])){
$username = $_POST['stdusername'];
$stdpass = md5($_POST['stdpassword']);
$stdemail = $_POST['stdemail'];
$connection = mysqli_connect("localhost","root","","std");
$query = "UPDATE student SET password='$stdpass', email='$stdemail' WHERE username LIKE '$username'";
mysqli_query($connection,$query);
echo mysqli_error($connection);
mysqli_close($connection);
}
if ( isset($_POST['stdsearch']) ){
$searchUsername = $_POST['stdeditusername'];
$connection = mysqli_connect("localhost","root","","std");
$query = "SELECT * FROM student WHERE username LIKE '$searchUsername' ";
$result = mysqli_query($connection, $query);
while( $row = mysqli_fetch_array($result) ){
$searchedUsername = $row['username'];
$searchedEmail = $row['email'];
}
}
?>
<html>
<head>
</head>
<body>
<h1>Student Registration</h1>
<form name="stdregistration" action="forms.php" method="post">
<label>Username :</label>
<input name="stdusername" required="required" type="text" /><br /><br />
<label>Password :</label>
<input name="stdpassword" type="password" /><br /><br />
<label>E-mail :</label>
<input name="stdemail" type="email" /><br /><br />
<input name="stdregister" type="submit" value="Save" />
</form>
<h2>Delete Students</h2>
<form name="stddeletion" action="forms.php" method="post">
<label>Select the Username :</label>
<select name="stddelusername" required>
<option value="">Select One</option>
<?php
$connection2 = mysqli_connect("localhost","root","","std");
$query2 = "SELECT username FROM student";
$result = mysqli_query($connection2,$query2);
while( $row = mysqli_fetch_array($result) ){
echo "<option value='".$row['username']."'>".$row['username']."</option>";
}
mysqli_close($connection2);
?>
</select>
<input name="stddelete" type="submit" value="Delete" />
</form>
<h2>Edit Students</h2>
<form name="stdedition" action="forms.php" method="post">
<label>Select the Username :</label>
<select name="stdeditusername" required>
<option value="">Select One</option>
<?php
$connection2 = mysqli_connect("localhost","root","","std");
$query2 = "SELECT username FROM student";
$result = mysqli_query($connection2,$query2);
while( $row = mysqli_fetch_array($result) ){
echo "<option value='".$row['username']."'>".$row['username']."</option>";
}
mysqli_close($connection2);
?>
</select>
<input name="stdsearch" type="submit" value="Search" />
</form>
<form name="stdedit" action="forms.php" method="post">
<label>Username :</label>
<input name="stdusername" required="required" type="text" readonly value="<?php echo $searchedUsername; ?>" /><br /><br />
<label>Password :</label>
<input name="stdpassword" type="password" /><br /><br />
<label>E-mail :</label>
<input name="stdemail" type="email" value="<?php echo $searchedEmail; ?>" /><br /><br />
<input name="stdupdate" type="submit" value="Update" />
</form>
</body>
</html>
回答by Dale
I recommend doing this in studentEdit.php
我建议在 studentEdit.php 中这样做
$student_id = mysql_real_escape_string($_GET[id]);
$query = "SELECT * FROM students WHERE id = '$student_id'";
$result = mysql_query($query) or die(mysql_error() . ' ' . $query);
$row = mysql_fetch_array($result);
mysql_close();
Two things I've changed here is firstly to escape the data being passed in the url and secondly I've added or die(mysql_error() . ' ' . $query);If something is going wrong in the sql statement you should now see the error and hopefully you'll be able to fix it from there.
我在这里更改的两件事首先是转义在 url 中传递的数据，其次我添加了or die(mysql_error() . ' ' . $query);如果 sql 语句中出现问题，您现在应该看到错误，希望您能够从那里。
回答by Conrad Lotz
What looks incorrect to me is the way you are displaying the value retrieved from the database:
在我看来不正确的是您显示从数据库中检索到的值的方式：
<input type="hidden" name="id" value="<? echo "$row[id]" ?>">
It should be
它应该是
<input type="hidden" name="id" value="<?php echo $row['id']; ?>">
