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php - 尝试插入/更新具有多个值的两个不同表

转载 作者:行者123 更新时间:2023-11-29 00:13:02 25 4
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我正在尝试使用 PHP 代码更新 mySQL 数据库中的两个不同表。一个 block 工作完美,但从“if(isset ...”我得到“错误查询数据库”消息,所以很明显我试图添加数组值(复选框)的代码不起作用。我不是什么看到了吗?

编辑.PHP

   <h3>Edit Profile: <?php echo $_REQUEST['first_name'];?></h3>
<form enctype="multipart/form-data" method="POST" action="change.php">
<table border="0" width="60%">

<tr><td width="30%">First Name: </td><td><input type="text"
name="upd_first_name" value="<?php echo $_REQUEST['first_name'];?>" maxlength="20"> </td></tr>

<tr><td width="30%">Last Name: </td><td><input type="text"
name="upd_last_name" value="<?php echo $_REQUEST['last_name'];?>" maxlength="20"> </td></tr>

<tr><td width="30%">Email: </td><td><input type="text"
name="upd_email" value="<?php echo $_REQUEST['email'];?>" maxlength="45"> </td></tr>

<tr><td width="30%">Password: </td><td><input type="password"
name="upd_password" id="upd_password" value="<?php echo $_REQUEST['password'];?>" maxlength="20"> </td></tr>

<tr><td width="30%">Confirm Password: </td><td><input type="password"
name="upd_cpassword" id="upd_cpassword" value="<?php echo $_REQUEST['password'];?>" maxlength="20" onkeyup="checkPass(); return false;"> </td></tr>

<tr><td width="30%">Profile Visbility: </td><td><input type="radio" name="upd_profilevis" value="1" id="1" checked> Private <input type="radio" name="upd_profilevis" value="2" id="2" > Public </td></tr>

<<tr><td width="30%">Industries: </td><td>
<input type="checkbox" name="industries[]" value="1"/>None</br>
<input type="checkbox" name="industries[]" value="2"/>Film</br>
<input type="checkbox" name="industries[]" value="3"/>Television</br>
<input type="checkbox" name="industries[]" value="4"/>Music</br>
<input type="checkbox" name="industries[]" value="5"/>Gaming</br>
<input type="checkbox" name="industries[]" value="6"/>Books</br>
<input type="checkbox" name="industries[]" value="7"/>Comic Books</br>
</td></tr>

<tr><td width="30%">Link: </td><td><input type="text"
name="upd_link" value="<?php echo $_REQUEST['profile_link'];?>" maxlength="45"> </td></tr>

<tr><td width="30%">Bio: </td><td><input type="text"
name="upd_bio" value="<?php echo $_REQUEST['bio'];?>" maxlength="500"> </td></tr>

<input type="hidden" name="MAX_FILE_SIZE" value="10000000">
<tr><td width="30%">Picture: </td><td><input type="file" id="image" name="image"></tr>
</table>
<span id="confirmMessage" class="confirmMessage"></span><br />

<input type="submit" value="Save & Update"/>
<input type="hidden" name="id" value="<?php echo $_REQUEST['id'];?>">
</form>

CHANGE.PHP(我得到错误的地方)

<?php
//pulls data/files from the edit.php form
$id= $_REQUEST['id'];
$upd_first_name= $_REQUEST['upd_first_name'];
$upd_last_name= $_REQUEST['upd_last_name'];
$upd_email= $_REQUEST['upd_email'];
$upd_password= $_REQUEST['upd_password'];
$upd_cpassword= $_REQUEST['upd_cpassword'];
$upd_profilevis= $_REQUEST['upd_profilevis'];
$upd_link= $_REQUEST['upd_link'];
$upd_bio= $_REQUEST['upd_bio'];
$mypic = $_FILES['image']['name'];
$temp = $_FILES['image']['tmp_name'];
$type = $_FILES['image']['type'];


//checks if the picture is of the right type before inserting
if(($type=="image/jpeg") || ($type=="image/jpg") || ($type=="image/png") || ($type=="image/bmp") || ($type=="image/gif")) {

//connects to the database
$dbc = mysqli_connect('localhost', 'root', 'root', 'profile') or die('Error connecting to MySQL server.');

//if industries are selected on the form edit.php insert into table
//each industry assigned to the member ID (junction table)
if(isset($_POST['industries'])) {

//deletes from table current values
$query = "DELETE FROM `industry_has_member` WHERE member_idmember='$id'";
$result = mysqli_query($dbc, $query) or die(mysqli_error($dbc));
//to report errors
if(!$result)
{
printf("Errormessage: %s\n", mysqli_error($dbc));
}
//adds on table new values
foreach ($_POST['industries'] as $industry) {
$query = "INSERT INTO industry_has_member (industry_idindustry, member_idmember) values ('$industry', '$id')";
$result = mysqli_query($dbc, $query) or die(mysqli_error($dbc));
//to report errors
if(!$result)
{
printf("Errormessage: %s\n", mysqli_error($dbc));
}
}
}

$query = "UPDATE `member` SET member_stamp=now(), first_name='$upd_first_name', last_name='$upd_last_name', email='$upd_email', password='$upd_password', profile_vis_idprofile_vis='$upd_profilevis', profile_link='$upd_link', prof_image='$mypic', bio='$upd_bio' WHERE idmember='$id'";

//displays the results into a table
$result = mysqli_query($dbc, $query) or die('Error querying database.');

move_uploaded_file($temp,"images/$mypic");

echo "<h2>Your profile has been updated</h2><br /><h3>Here is your profile picture</h3>";
echo "<img border='1' width='200' height='200' src='images/$mypic'>";

mysqli_close($dbc);

}
?>

最佳答案

您需要将所有操作包装在一个事务中。

if (isset($_POST['industries'])) {
$industry_ids = $_POST['industries'];
$id = $_POST['id'];

try {
$dbc->begin_transaction();
$dbc->query("delete from industry_has_member where member_idmember=$id") or throw new Exception($dbc->error);

foreach ($industry_ids as $industry_id) {
$dbc->query("insert ignore into industry_has_member (industry_idindustry, member_idmember) values ($industry_id, $id)" or throw new Exception($dbc->error);
}
$dbc->commit();
} catch (Exception $e) {
echo $e->getMessage();
$dbc->rollback();
}
}

关于php - 尝试插入/更新具有多个值的两个不同表,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/24100041/

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