錯誤的SQL查詢php

Question

我得到這個錯誤，當我嘗試用som moviegenres更新我的數據庫。

UPDATE RM_Movie2Genre SET 

Error in executing query: 42000 Array 
(
[0] => 42000 
[1] => 1064 
[2] => You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 1 

電影被保存但不與流派。任何人都可以看到這裏有什麼問題？我知道這是在UPDATE RM_Movie2Genre SET（標記爲粗體），但我看不出有什麼不對......

$output = null; 
if($save) { 
// First save information on movie 
$sql = ' 
UPDATE RM_Movie SET 
    title = ?, 
    year = ?, 
    image = ?, 
    pris = ?, 
    director = ?, 
    length = ?, 
    plot = ?, 
    imdb = ?, 
    trailer = ?, 
    filter = ? 
WHERE 
    id = ? 
'; 
$params = array($title, $year, $image, $pris, $director, $length, $plot, $imdb, $trailer, $filter, $id); 
$db->ExecuteQuery($sql, $params); 

// Then save information on genre 
foreach($genre as $val) { 
    **$db->ExecuteQuery('UPDATE RM_Movie2Genre SET', array($val));** 
} 
$output = 'Informationen sparades'; 
} 

此代碼將只能與更新/插入電影替換所有電影。我將使用我的代碼進行更新和插入。

// Get all genres and select those that should be selected 
$sql = 'SELECT * FROM RM_MovieGenre'; 
$genres = $db->ExecuteSelectQueryAndFetchAll($sql); 

$sql = 'SELECT idRM_MovieGenre AS id FROM RM_Movie2Genre WHERE idRM_Movie = ?'; 
$myGenres = $db->ExecuteSelectQueryAndFetchAll($sql, array($id), 0, PDO::FETCH_COLUMN); 

$selectOptionGenres = "<select multiple name='genre[]'>"; 
foreach($genres as $key => $val) { 
$selected = in_array($val->id, $myGenres) ? ' selected' : null; 
$selectOptionGenres .= "<option value='{$val->id}'{$selected}>{$val->name}</option>"; 
} 
$selectOptionGenres .= '</select>'; 


// Check if form was submitted 
$output = null; 
if($save) { 
// First save information on movie 
$sql = ' 
UPDATE RM_Movie SET 
    title = ?, 
    year = ?, 
    image = ?, 
    pris = ?, 
    director = ?, 
    length = ?, 
    plot = ?, 
    imdb = ?, 
    trailer = ?, 
    filter = ? 
'; 
$params = array($title, $year, $image, $pris, $director, $length, $plot, $imdb, $trailer, $filter); 
$db->ExecuteQuery($sql, $params); 

// Then save information on genre 
$query="INSERT INTO RM_Movie2Genre(idRM_Movie,idRM_MovieGenre) values($id,?)"; 

foreach($genre as $val) { 
    $val_array=array($val); 
    $db->ExecuteQuery($query, $val_array); 
} 
$output = 'Informationen sparades'; 
} 

// Select information on the movie 
$sql = 'SELECT * FROM RM_Movie WHERE id = ?'; 
$params = array($id); 
$res = $db->ExecuteSelectQueryAndFetchAll($sql, $params); 

if(isset($res[0])) { 
$movie = $res[0]; 
} 
else { 
die('Failed: There is no movie with that id'); 
} 

Answer 1

我想你想改變的行與價值：

$id=$params[10]; 

我認爲你有一個ID多個流派，你把它存儲在RM_Movie2Genre表。如果是這種情況，你應該使用INSERT查詢。但首先，你應該刪除舊值：

$query="DELETE FROM RM_Movie2Genre where id=?"; 

    $val_array=array($id); 
    $db->ExecuteQuery($query, $val_array); 

現在，用於插入，如果$ id爲數字使用此查詢：

$query="INSERT INTO RM_Movie2Genre(id,genre) values($id,?)"; 

否則，如果它是字符串類型的，你應該換$ ID用單引號：

$query="INSERT INTO RM_Movie2Genre(id,genre) values('$id',?)"; 

最後：

foreach($genre as $val) { 
    $val_array=array($val); 
    $db->ExecuteQuery($query, $val_array); 
} 

我不認爲UPDATE是有意義的，因爲使用更新查詢，您將多次更新單行，這是無用的。無論如何，如果你仍然認爲update查詢，那麼你應該與這一個改變$query：

$query="UPDATE RM_Movie2Genre SET genre = ? WHERE id =$id"; 

，如果它是字符串類型：

$query="UPDATE RM_Movie2Genre SET genre = ? WHERE id ='$id'"; 

Answer 2

此代碼的工作只是罰款，既創建和編輯。謝謝你讓我到達那裏！ :-)

// Get all genres and select those that should be selected 
$sql = 'SELECT * FROM RM_MovieGenre'; 
$genres = $db->ExecuteSelectQueryAndFetchAll($sql); 

$sql = 'SELECT idRM_MovieGenre AS id FROM RM_Movie2Genre WHERE idRM_Movie = ?'; 
$myGenres = $db->ExecuteSelectQueryAndFetchAll($sql, array($id), 0, PDO::FETCH_COLUMN); 

$selectOptionGenres = "<select multiple name='genre[]'>"; 
foreach($genres as $key => $val) { 
    $selected = in_array($val->id, $myGenres) ? ' selected' : null; 
    $selectOptionGenres .= "<option value='{$val->id}'{$selected}>{$val->name}</option>"; 
} 
$selectOptionGenres .= '</select>'; 


// Check if form was submitted 
$output = null; 
if($save) { 
    // First save information on movie 
    $sql = ' 
    UPDATE RM_Movie SET 
     title = ?, 
     year = ?, 
     image = ?, 
     pris = ?, 
     director = ?, 
     length = ?, 
     plot = ?, 
     imdb = ?, 
     trailer = ?, 
     filter = ? 
    WHERE 
     id = ? 
    '; 
    $params = array($title, $year, $image, $pris, $director, $length, $plot, $imdb, $trailer, $filter, $id); 
    $db->ExecuteQuery($sql, $params); 

    // Then save information on genre 
    $query="INSERT INTO RM_Movie2Genre(idRM_Movie,idRM_MovieGenre) values($id,?)"; 

    foreach($genre as $val) { 
     $val_array=array($val); 
     $db->ExecuteQuery($query, $val_array); 
    } 
    $output = 'Informationen sparades'; 
} 

// Select information on the movie 
$sql = 'SELECT * FROM RM_Movie WHERE id = ?'; 
$params = array($id); 
$res = $db->ExecuteSelectQueryAndFetchAll($sql, $params); 

if(isset($res[0])) { 
    $movie = $res[0]; 
} 
else { 
    die('Failed: There is no movie with that id'); 
}