php获取用户下拉列表选择
我如何为每个用户选择下拉菜单。
user table
------------
id job
1 1
2 2
job table
----------
id name
1 Doctor
2 Sales
$q = $db->query("SELECT * FROM affiliate LEFT JOIN user ON user.job = affiliate.id_affiliate");
while($r = $q->fetch_array()) :
if($r['id_user'] == $_SESSION['id_user'] && $r['job'] == $r['id_affiliate']) {
echo '<option selected value="'.$r['id_affiliate'].'">'.$r['org'].'</option>';
} else {
echo '<option value="'.$r['id_affiliate'].'">'.$r['org'].'</option>';
}
endwhile;
how do i get dropdown selected for each user.
user table
------------
id job
1 1
2 2
job table
----------
id name
1 Doctor
2 Sales
$q = $db->query("SELECT * FROM affiliate LEFT JOIN user ON user.job = affiliate.id_affiliate");
while($r = $q->fetch_array()) :
if($r['id_user'] == $_SESSION['id_user'] && $r['job'] == $r['id_affiliate']) {
echo '<option selected value="'.$r['id_affiliate'].'">'.$r['org'].'</option>';
} else {
echo '<option value="'.$r['id_affiliate'].'">'.$r['org'].'</option>';
}
endwhile;
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selected="selected"或只是selected应该正常工作。如果不是，则说明您的 if 语句有问题。一种简单的方法是回显 if 语句的内容，如下所示：注意！回显通常应该在选择开放标签之外完成，只需将以下内容粘贴到选择开放标签之外，但就在查询之后。
您现在可以检查这些值是否确实匹配。如果没有，那就是你的问题。
selected="selected"or justselectedshould normally work. If not there is a problem with your if statement. One simple way is to echo out the content of the if statement like this:note!! the echo should normally be done outside the select open tag, just paste the following outside the select open tag but just after your query.
you can now check if the values actually match. if not then there is your problem.
修改以下内容怎么样...
不确定这是否重要，但我在选项末尾选择了
选择。How about modifying the following ...
Not sure if it matters, but I have
selectedat the end of my option.