[英]PHP/MySQL outputting specified fields based upon previous if statements
[英]PHP Echo Records Based Upon specified IDs
所以我需要一种方法,可以在指定了ID的地方echo $row['username']
和echo $row['password']
。
例如,我有3行,我需要从ID号$query = "SELECT * FROM Card_File WHERE id = $id";
行中获取信息。现在,我进行了查询$query = "SELECT * FROM Card_File WHERE id = $id";
我该如何回显这些数据(请记住,它不在表格中,并且可以放置在网页上的随机位置,并且输出将进入的表格将有所不同(例如,没有相同的表格字段名称)。 )
echo $row['username'] $id=20;
echo $row['password'] $id=20;
echo $row['username'] $id=56;
echo $row['password'] $id=56;
echo $row['username'] $id=88;
echo $row['password'] $id=88;
我一直在寻找周围的事物,并看到了一些非常复杂的示例,但不要相信它很难被id逐行回显。 我可能是错的
下面不是最终结果的一个很好的例子,但是可以给您一个想法。
<!DOCTYPE html> <html> <head> <?php // Connect to the database $link = mysqli_connect('localhost','root','pass','ACS_SHAREPOINT_ADDONS'); // Select table and arguments $query = "SELECT * FROM Card_File WHERE id = $id"; $result = mysqli_query($link, $query); ?> </head> <body> <?php foreach($row = mysqli_fetch_array($result)): ?> <div class="loginForms"> <form target="_blank" action="https://www.website.com" method="post"> <input type="hidden" name="username" id="auth-form-login" value="<?php $id=20 echo $row['username']; ?>"/> <input type="hidden" name="password" id="auth-form-pass" value="<?php $id=20 echo $row['password']; ?>"/> <input type="hidden" name="redirDocument" value="user"/> <input type="hidden" name="query" value=""/> <input type="hidden" name="urlString" value="doc-rc-login;lng-ww-en;tpl-;ver-;"/> <input type="submit" name="authSubmit" value="AVG Resellers Center" id="auth-form-submit"/> </form> </div> <div class="loginForms"> <form target="_blank" action="https://www.website.com" method="post"> <input type="hidden" name="username" id="auth-form-login" value="<?php $id=56 echo $row['username']; ?>"/> <input type="hidden" name="password" id="auth-form-pass" value="<?php $id=56 echo $row['password']; ?>"/> <input type="hidden" name="redirDocument" value="user"/> <input type="hidden" name="query" value=""/> <input type="hidden" name="urlString" value="doc-rc-login;lng-ww-en;tpl-;ver-;"/> <input type="submit" name="authSubmit" value="AVG Resellers Center" id="auth-form-submit"/> </form> </div> <div class="loginForms"> <form target="_blank" action="https://www.website.com" method="post"> <input type="hidden" name="username" id="auth-form-login" value="<?php $id=88 echo $row['username']; ?>"/> <input type="hidden" name="password" id="auth-form-pass" value="<?php $id=88 echo $row['password']; ?>"/> <input type="hidden" name="redirDocument" value="user"/> <input type="hidden" name="query" value=""/> <input type="hidden" name="urlString" value="doc-rc-login;lng-ww-en;tpl-;ver-;"/> <input type="submit" name="authSubmit" value="AVG Resellers Center" id="auth-form-submit"/> </form> </div> <?php endforeach; ?> </body> </html>
我相信这是您想要做的。
<?php
$ids_tofind = array(1,6,3); //for example
$id = implode(",",$ids_tofind);
$link = mysqli_connect('localhost','root','pass','ACS_SHAREPOINT_ADDONS');
// Select table and arguments
$query = "SELECT * FROM Card_File WHERE id IN(". $id.")";
$result = mysqli_query($link, $query);
while($row = mysqli_fetch_array($result)){;?>
<div class="loginForms">
<form target="_blank" action="https://www.website.com" method="post">
<input type="hidden" name="username" id="auth-form-login" value="<?php echo $row['username']; ?>"/>
<input type="hidden" name="password" id="auth-form-pass" value="<?php echo $row['password']; ?>"/>
<input type="hidden" name="redirDocument" value="user"/>
<input type="hidden" name="query" value=""/>
<input type="hidden" name="urlString" value="doc-rc-login;lng-ww-en;tpl-;ver-;"/>
<input type="submit" name="authSubmit" value="AVG Resellers Center" id="auth-form-submit"/>
</form>
</div>
<?php };?>
这将回显ID为1、6和3的所有三个用户的表单。
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