Question

我正在研究MySQL＆amp; PHP项目。它基于音乐数据库。当我转到http://andrewb1.sgedu.site/editgenres.php时，我收到以下错误：

Error: SQL Error: 
Errno: 1064
Error: 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

我有点困惑我如何在第1行收到错误，因为唯一有开放的php标记

editgenres.php的代码是：

<?php

include 'dbconnect.php';

$sql = "select * from Genres where GenreID = " . $_REQUEST['GenreID'];
if (!$result = $mysqli->query($sql)) {
    echo "Error: SQL Error: </br>";
    echo "Errno: " . $mysqli->errno . "</br>";
    echo "Error: " . $mysqli->error . "</br>";

    exit;
}

$row = $result->fetch_assoc();

?>

<form action="editgenressrv.php">
<input type="hidden" name="GenreID" value = "<?php echo $row["GenreID"]?>"/> 
GenreID:<input type="text" name="GenreID" value="<?php echo $row["GenreID"]?>"/></br>
GenreName:<input type="text" name="GenreName" value="<?php echo $row["GenreName"]?>"/></br>
<input type="submit"/>
</form>

此外，如果需要，这里是EditGenresSrv.php的代码：

include 'dbconnect.php';

$sql = "update Genres set ";
$sql .= "GenreID = '" . $_REQUEST["firstname"] ."'," ;
$sql .= "GenreName = '" . $_REQUEST["lastname"] ."'," ;
$sql .= "where GenreID= " . $_REQUEST['GenreID']; 
if (!$result = $mysqli->query($sql)) {
    echo "Error: SQL Error: </br>";
    echo "Errno: " . $mysqli->errno . "</br>";
    echo "Error: " . $mysqli->error . "</br>";

    exit;
}
?>
<script>
window.location='genres.php';
</script>

如果需要，这里是dbconnect.php（虽然我已经测试了它并且很好）：

include 'dbconnect.php';

$sql = "insert into students (firstname,lastname,email) values (" . 
  "'" . $_REQUEST["GenreID"] ."','" .
  $_REQUEST["GenreName"] . "' ";

if (!$result = $mysqli->query($sql)) {
    echo "Error: SQL Error: </br>";
    echo "Errno: " . $mysqli->errno . "</br>";
    echo "Error: " . $mysqli->error . "</br>";

    exit;
}
?>
<script>
window.location='genres.php';
</script>

这是HTM文件：

<form action="addgenressrv.php">
GenreID:<input type="text" name="GenreID"/></br>
GenreName:<input type="text" name="GenreName"/></br>

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

Answer 1

在逗号之前小心使用逗号。

$sql = "update Genres set ";
$sql .= "GenreID = '" . $_REQUEST["firstname"] ."'," ;
$sql .= "GenreName = '" . $_REQUEST["lastname"] ."' " ;
$sql .= "where GenreID= " . $_REQUEST['GenreID'];

Answer 2

您需要将GenreID传递到您的页面，请查看以下链接

http://andrewb1.sgedu.site/editgenres.php?GenreID=1

你会明白一切。如果没有，那么我会解释你。您的上一页应该有$_REQUEST['GenreID']的值。

$sql = "select * from Genres where GenreID = " . $_REQUEST['GenreID'];

这一行给出了错误消息，因为无论是使用POST方法还是GET，您都没有将GenreID传递给文件editgenres.php。

在表单中添加<form action="editgenres.php">然后

GenreID:<input type="text" name="GenreID"/></br>
GenreName:<input type="text" name="GenreName"/></br>

因为当你用editgenres.php说错误时，你必须按照上面的表格调用此页面。检查您的第一页的操作，该页面将调用http://andrewb1.sgedu.site/editgenres.php

MySQL错误1064：您的SQL语法有错误..在第1行附近？

2 个答案: