Php 从浏览器而不是编辑器获取SQL语法错误

我试图编写一个sql语法，但它一直给我一个错误，即使dreamweaver说没有错误。如有任何反馈，将不胜感激

这就是我所拥有的：

$inner_result = mysql_query("SELECT * FROM product WHERE manID = ".$row['id']." ") or die(mysql_error());

我曾尝试在第二组引号前面加反斜杠，但没有成功

我也试过这个

('SELECT * FROM product WHERE manID = ".$row['id']." ORDER BY name') 

('SELECT * FROM product WHERE manID = '.$row['id'].' ') 

('SELECT * FROM product WHERE manID = ".$row['id']." ')

("SELECT * FROM product WHERE manID = '.$row['id'].' ")

("SELECT * FROM product WHERE manID = '.$row["id"].' ")

但还是给了我错误

您的SQL语法有错误；检查与MySQL服务器版本对应的手册，以了解第1行“ORDER BY name ASC”附近使用的正确语法

我也试过这个

('SELECT * FROM product WHERE manID = ".$row['id']." ORDER BY name') 

('SELECT * FROM product WHERE manID = '.$row['id'].' ') 

('SELECT * FROM product WHERE manID = ".$row['id']." ')

("SELECT * FROM product WHERE manID = '.$row['id'].' ")

("SELECT * FROM product WHERE manID = '.$row["id"].' ")

但还是给了我错误。这是我在另一个页面上使用的，它正在工作。我复制粘贴了准确的代码，现在它给了我一个错误

我也试过这个

('SELECT * FROM product WHERE manID = ".$row['id']." ORDER BY name') 

('SELECT * FROM product WHERE manID = '.$row['id'].' ') 

('SELECT * FROM product WHERE manID = ".$row['id']." ')

("SELECT * FROM product WHERE manID = '.$row['id'].' ")

("SELECT * FROM product WHERE manID = '.$row["id"].' ")

我也试过这个

('SELECT * FROM product WHERE manID = ".$row['id']." ORDER BY name') 

('SELECT * FROM product WHERE manID = '.$row['id'].' ') 

('SELECT * FROM product WHERE manID = ".$row['id']." ')

("SELECT * FROM product WHERE manID = '.$row['id'].' ")

("SELECT * FROM product WHERE manID = '.$row["id"].' ")

我也试过这个

('SELECT * FROM product WHERE manID = ".$row['id']." ORDER BY name') 

('SELECT * FROM product WHERE manID = '.$row['id'].' ') 

('SELECT * FROM product WHERE manID = ".$row['id']." ')

("SELECT * FROM product WHERE manID = '.$row['id'].' ")

("SELECT * FROM product WHERE manID = '.$row["id"].' ")

当然，在尝试上述方法时，dreamweaver告诉我存在语法错误

以下是代码的一部分：

$inner_result = mysql_query("SELECT * FROM product WHERE manID = ".$row['id']." ORDER BY name ASC") or die(mysql_error());  
if(mysql_num_rows($inner_result) > 0){
    echo '  </div>
        <ul>';
    while($inner_row = mysql_fetch_array($inner_result)) {
    ?>
    <li>
        <div>
            <span class="row-name"><?= $inner_row['name'] ?></span>
        </div>
    </li>
    <?php
        }
        echo '</ul>';
        }
    ?>

$internal\u result=mysql\u查询（“从产品中选择*，其中manID=“.$row['id']”按名称排序ASC”）或die（mysql\u error（））；
如果（mysql_num_rows（$internal_result）>0）{
回声'
”；
而（$inner\u row=mysql\u fetch\u数组（$inner\u result））{
?>

试着这样做，在

=

"SELECT * FROM product WHERE manID = ".$row['id']." ORDER BY name"

---

什么是“它”这给了您一个错误，错误消息是什么？对不起！应该已经提到了。浏览器给我的错误是：您的SQL语法有错误；请查看与您的MySQL服务器版本对应的手册，以获得在第1行“ORDER BY name ASC”附近使用的正确语法，以便调试：您可以尝试并echo“SELECT*FROM product WHERE manID=“.$row['id']”按名称订购ASC”然后将其传递给函数并查看输出？可能是

$row['id']

为空或包含非数字字符。它不是空的，我在CMS部分中使用了相同的代码，并且正在显示。同时在我的数据库中填充了列。没有。仍然是相同的错误。如果我把整个代码放上去，会有帮助吗？我放了编辑器允许我放的代码。

$row['id'的值是多少

？它是每个产品的实际id，我已将其设置为自动递增。它是一个INT。希望这能有所帮助。您能

回送

要查看的查询；当查询到达DB时它的外观如何？