我正在尝试为我自己的国家制作this site的镜像,但我很难理解如何从我的数据库条目中创建新的网址。
例如,在我链接的网站上,它有和id / name来识别那个确切的网站,我想达到同样的目的,使单独的问题可以共享。
我的当前代码仅用于在localhost网址上测试1个问题。
这是我目前得到的一小段内容:
<?php
$id = 1;
$result = performQuery("SELECT * FROM dilemmas WHERE id = ".$id);
while($row = mysqli_fetch_array($result)){
$blue_question = utf8_encode($row['blue_question']);
$blue_votes = utf8_encode($row['blue_votes']);
$red_question = utf8_encode($row['red_question']);
$red_votes = utf8_encode($row['red_votes']);
}
?>
<div class="answer-peek">
<div class="peek-text">Vil du helst...</div>
<div class="peek-buttons">
<div class="peek-bluebutton"><?php echo $blue_question ?></div>
<div class="peek-redbutton"><?php echo $red_question ?></div>
</div>
</div>
<div class="row text-center" id="textrow">
<h3 style="color:white;">Vil du helst...</h3>
</div>
<div class="row" id="buttonrow">
<div class="col-md-1">
<div class="button-left"></div>
</div>
<div class="col-md-10"><
<div class="col-md-5 col-md-offset-1 button" id="bluebutton">
<div class="blueCheckDiv"></div>
<div id="text-container">
<table cellpadding="0">
<tbody>
<tr>
<td valign="middle">
<div class="result">
<div class="percentage">
<?php
$percent = $blue_votes/($blue_votes + $red_votes);
$percent_friendly = number_format( $percent * 100, 0 ) . '%';
echo $percent_friendly;
?>
</div>
<div class="total-votes">
<span class="count">
<?php echo $blue_votes; ?>
</span>
<span class="word">
enige
</span>
</div>
<div class="question-text">
<?php echo $blue_question ?>
</div>
</div>
<p class="question">
<?php echo $blue_question ?>
</p>
</td>
</tr>
</tbody>
</table>
</div>
</div>
<div class="col-md-5 button" id="redbutton">
<div class="redCheckDiv"></div>
<div id="text-container">
<table cellpadding="0">
<tbody>
<tr>
<td valign="middle">
<div class="result">
<div class="percentage">
<?php
$percent = $blue_votes/($blue_votes + $red_votes);
$percent_friendly = number_format( $percent * 100, 2 ) . '%';
echo $percent_friendly;
?>
</div>
<div class="total-votes">
<span class="count">
<?php echo $red_votes ?>
</span>
<span class="word">
uenige
</span>
</div>
<div class="question-text">
<?php echo $red_question ?>
</div>
</div>
<p class="question">
<?php echo $red_question ?>
</p>
</td>
</tr>
</tbody>
</table>
</div>
</div>
</div>
<div class="col-md-1">
<div class="button-right"></div>
</div>
</div>
<div class="row" id="dividerrow">
<div class="col-md-8" id="divider-left"></div>
<div class="col-md-4" id="divider-right"></div>
</div>
<div class="row" id="socialrow">
<div class="col-md-8 socialcol" id="col-left"></div>
<div class="col-md-4 socialcol" id="col-right"></div>
</div>
答案 0 :(得分:0)
我建议你在评论的评论中使用一个框架。
无论如何,要做你正在尝试的事情,你可以尝试这样的事情:
<?php
$id = 1;
if (true === isset($_GET['id'])) {
$id = $_GET['id'];
}
$result = performQuery("SELECT * FROM dilemmas WHERE id = ".$id);
while($row = mysqli_fetch_array($result)){
$blue_question = utf8_encode($row['blue_question']);
$blue_votes = utf8_encode($row['blue_votes']);
$red_question = utf8_encode($row['red_question']);
$red_votes = utf8_encode($row['red_votes']);
}
您可以使用example.com/page.php?id=1234
。
因此,您可以为每个问题构建链接:
<a href="/page.php?id=<?php echo $row['id']; ?>">Share</a>
注意:这是一个非常脏的代码,它会导致许多安全问题。我写它只是为了让你能够理解动态,但是,我建议你在第一次开始时使用一个框架:make all in&#34; plain&#34; PHP将使您处理许多复杂的问题,如安全性或路由或服务的处理等...使用框架找到非常好的指导,并更快更好地学习您选择的语言(PHP在此情况)。