dongshi1148 2015-12-11 00:13
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PHP回送中PHP文件的href链接不起作用

I have created a table called menu and two of its columns are menu_option and menu_content. What I want to do is, after I click on a menu_option to go to a separate .php file i created, which is supposed to show in a div the menu_content.

I still haven't figured out why when I click on it, it tells me that the object cannot be found.

<?php
$res=mysql_query("SELECT * FROM menu");
while($row=mysql_fetch_array($res)){
    echo "<ul>";
        echo "<li>";    
            echo "<a href=\"menucontent.php\">";
            echo $row['menu_option']; 
            echo '</a>';
        echo "</li>";
    echo "</ul>";
}
?>

At first I thought this could work and in my menucontent.php I could use echo $row['menu_content']; to print the proper menu content for the menu option.

I would really apreciate any suggestions.

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1条回答 默认 最新

  • doutan8775 2016-02-18 12:37
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    I have figured it out. Thank you all for the suggestions. I simply deleted the existing menucontent.php and created a new one. It worked. I replaced the a href link with '' where $idaux = $row['menu_id']. In my new file menuc.php i loop through the rows and when $row['menu_id'] === $_GET['id'] , i simply echo $row['menu_content'], which is what i wanted.

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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