以PHP格式编辑MYSQL行

Question

我以前做过这个,出于某种原因我这次不能让它工作!我把头发拉过来!因为没有错误,所以它不会更新数据库.

基本上我有一个学生数据表....

ID | IMGNU | 名字| 姓氏| FBID

我们以第233行为例.

我可以通过view.php查看特定的行？ID = 233

然后这可行,但现在我想能够去edit.php？ID = 233并且它应该加载一个表格,已经有来自行233的信息.然后我应该能够编辑字段中的数据和提交表单,这将改变数据库中的信息.

这就是我已经拥有的.

edit.php

<?php
mysql_connect('localhost', 'admin', 'passw0rd') or die(mysql_error());
echo "Tick <p>";
mysql_select_db("students") or die(mysql_error());
echo "Tick"; 

$UID = $_GET['ID'];

$query = mysql_query("SELECT * FROM stokesley_students WHERE id = '$UID'")
or die(mysql_error());  

while($row = mysql_fetch_array($query)) {
echo "";

$firstname = $row['firstname'];
$surname = $row['surname'];
$FBID = $row['FBID'];
$IMGNU = $row['IMGNU'];


};

?>

<form action="update.php?ID=<?php echo "$UID" ?>" method="post">

IMGNU: <input type="text" name="ud_img" value="<?php echo "$IMGNU" ?>"><br>

First Name: <input type="text" name="ud_firstname" value="<?php echo "$firstname" ?>"><br>

Last Name: <input type="text" name="ud_surname" value="<?php echo "$surname" ?>"><br>

FB: <input type="text" name="ud_FBID" value="<?php echo "$FBID" ?>"><br>

<input type="Submit">
</form>

这是update.php

<

?php

$ud_ID = $_GET["ID"];

$ud_firstname = $_POST["ud_firstname"];
$ud_surname = $_POST["ud_surname"];
$ud_FBID = $_POST["ud_FBID"];
$ud_IMG = $_POST["ud_IMG"];

mysql_connect('localhost', 'admin', 'passw0rd') or die(mysql_error());
echo "MySQL Connection Established! <br>";

mysql_select_db("students") or die(mysql_error());
echo "Database Found! <br>";


$query="UPDATE * stokesley_students SET firstname = '$ud_firstname', surname = '$ud_surname', 
FBID = '$ud_FBID' WHERE ID ='$ud_IMG'";

mysql_query($query);

echo "<p>Record Updated<p>";

mysql_close();
?>

任何想法都会非常感激,maby我只是错过了一些愚蠢的东西？

谢谢Alex

Answer 1

edit.php - 有一些变化

<?php
mysql_connect('localhost', 'admin', 'passw0rd') or die(mysql_error());
mysql_select_db("students") or die(mysql_error());

$UID = (int)$_GET['ID'];
$query = mysql_query("SELECT * FROM stokesley_students WHERE id = '$UID'") or die(mysql_error());

if(mysql_num_rows($query)>=1){
    while($row = mysql_fetch_array($query)) {
        $firstname = $row['firstname'];
        $surname = $row['surname'];
        $FBID = $row['FBID'];
        $IMGNU = $row['IMGNU'];
    }
?>
<form action="update.php" method="post">
<input type="hidden" name="ID" value="<?=$UID;?>">
IMGNU: <input type="text" name="ud_img" value="<?=$IMGNU;?>"><br>
First Name: <input type="text" name="ud_firstname" value="<?=$firstname?>"><br>
Last Name: <input type="text" name="ud_surname" value="<?=$surname?>"><br>
FB: <input type="text" name="ud_FBID" value="<?=$FBID?>"><br>
<input type="Submit">
</form>
<?php
}else{
    echo 'No entry found. <a href="javascript:history.back()">Go back</a>';
}
?>

update.php(除星号外,您的查询也ID与$ud_IMG变量匹配)

    <?php
    mysql_connect('localhost', 'admin', 'passw0rd') or die(mysql_error());
    mysql_select_db("students") or die(mysql_error());

    $ud_ID = (int)$_POST["ID"];

    $ud_firstname = mysql_real_escape_string($_POST["ud_firstname"]);
    $ud_surname = mysql_real_escape_string($_POST["ud_surname"]);
    $ud_FBID = mysql_real_escape_string($_POST["ud_FBID"]);
    $ud_IMG = mysql_real_escape_string($_POST["ud_IMG"]);


    $query="UPDATE stokesley_students
            SET firstname = '$ud_firstname', surname = '$ud_surname', FBID = '$ud_FBID' 
            WHERE ID='$ud_ID'";


mysql_query($query)or die(mysql_error());
if(mysql_affected_rows()>=1){
    echo "<p>($ud_ID) Record Updated<p>";
}else{
    echo "<p>($ud_ID) Not Updated<p>";
}
?>

Answer 2

"UPDATE * stokesley_students SET firstname = '$ud_firstname', surname = '$ud_surname', 
FBID = '$ud_FBID' WHERE ID ='$ud_IMG'"

那个查询是错误的，去掉星号。此外，您不知道是否有错误，因为您没有检查mysql_query 或使用mysql_error.