Question

我只是无法弄清楚为什么我收到错误消息，我已经尝试删除''和（）

我在phpmyadmin中运行了脚本，它说我的语法问题是$ result =（“SELECT * FROM'test_prefixCatagory'ORDDER by'Cat'”）;

$result = ("SELECT * FROM 'test_prefixCatagory' ORDER by 'Cat'");


while($row = mysql_fetch_array($result))

$sCat = ($row['Cat']);
$sCatID = ($row['CatID']);
{
echo "<table>";
 echo "<tr valign='top'><td><b><a href='#".$sCat."'>".$sCat."</a></b><br>";
 // column 1 categories
 $result2 = ("SELECT * FROM `test_prefixSubCat` WHERE `CatID`=$sCatID");
 // sub-cats
 while($row2 = mysql_fetch_array($result2)) 
    {
  $sSub = ($row2['CatID']);
  $sSubID = ($row2['SubID']);
  echo "<dd><a href='#'>".$sSub."</a><br>";

 }

 echo "<br></td></tr>";
echo "</table>";
 }

有人有想法吗？

Answer 1

试试这个：

<?php
$result = mysql_query("SELECT * FROM `test_prefixCatagory ORDER by `Cat`");

while ($row = mysql_fetch_array($result)) {
$sCat = $row['Cat'];
$sCatID = $row['CatID'];
echo "<table>";
echo "<tr valign='top'><td><b><a href='#" . $sCat . "'>" . $sCat . "</a></b><br>";
// column 1 categories
$result2 = mysql_query("SELECT * FROM `test_prefixSubCat` WHERE `CatID`='".$sCatID. "'");
// sub-cats
while ($row2 = mysql_fetch_array($result2)) {
    $sSub = $row2['CatID'];
    $sSubID = $row2['SubID'];
    echo "<dd><a href='#'>" . $sSub . "</a><br>";

    }

        echo "<br></td></tr>";
        echo "</table>";
}
?>

Answer 2

$result = ("SELECT * FROM `test_prefixCatagory` ORDER by `Cat`");

Answer 3

您不仅需要添加mysql_query，还需要从表名和字段名中删除单引号。如果您愿意，可以使用反引号，但不是表名周围的单引号。

$result = mysql_query("SELECT * FROM `test_prefixCatagory` ORDER by `Cat`");

// other query:
 $result2 = mysql_query("SELECT * FROM `test_prefixSubCat` WHERE `CatID`=$sCatID");

调试MySQL问题时，请使用mysql_error()查看问题说明。

SQL ERROR mysql_fetch_array（）：无效？

3 个答案: