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我不确定我的代码出了什么问题。我收到消息“页面无法正常工作”不知道为什么?我正在尝试获取“retrieve1.php”过滤器的详细信息,但每次尝试点击学生过滤器时,我都会收到一条消息,指出“页面无法正常工作”。我很感谢这里的任何帮助。我的php搜索代码有问题
<?php
echo "<body style='background-color:#DCDCDC'>";
include ("account.php");
($db = mysql_connect($hostname, $username, $password))
or die ("unable to connect to MYSQL database");
mysql_select_db($project);
$sql= "SELECT * FROM bpi_registration LEFT JOIN
bpi_schoolInfo on bpi_registration.id_school = bpi_schoolInfo.id_school";
\t
$query=mysql_query($sql) or die(mysql_error());
function grade()
{
\t $query= "select distinct class_name from bpi_classInfo";
\t $result=mysql_query($query) or die(mysql_error());
\t \t
\t \t
\t while ($value = mysql_fetch_array ($result))
\t {
\t echo "<option value='" . $value['class_name'] . "'>" . $value['class_name'] . "</option>";
\t }
}
function school()
{
\t $query= "select distinct school_name from bpi_schoolInfo";
\t $result=mysql_query($query)or die(mysql_error());
\t \t
\t \t
\t while ($value = mysql_fetch_array ($result))
\t {
\t \t echo "<option value='" . $value['school_name'] . "'>" . $value['school_name'] . "</option>";
\t }
}
function team()
{
\t $query= "select distinct team_name from bpi_teamProfile";
\t $result=mysql_query($query)or die(mysql_error());
\t \t
\t \t
\t while ($value = mysql_fetch_array ($result))
\t {
\t \t echo "<option value='" . $value['team_name'] . "'>" . $value['team_name'] . "</option>";
\t }
}
function students()
{
\t $query= "select * from bpi_registration";
\t $result=mysql_query($query)or die(mysql_error());
\t \t
\t \t
\t while ($value = mysql_fetch_array ($result))
\t {
\t \t echo "<option value='" . $value['first_name'].' '.$value['last_name']. "'>" . $value['first_name'].' '.$value['last_name']. "</option>";
\t \t
\t }
}
?>
<form action="retrieve1.php" method="GET">
<select name="Grade">
<option value="" selected="selected">Choose Grade</option>
<?php grade() ?>
</select>
<select name="School">
<option value="" selected="selected">Choose School</option>
<?php school() ?>
</select>
<select name="Team">
<option value="" selected="selected">Choose Team</option>
<?php team() ?>
</select>
<select name="Students">
<option value="" selected="selected">Choose Students</option>
<?php students() ?>
</select>
<input type="submit" value="Find" />
</form>
<table width="600" border="2">
<tr>
<th width="91"> <div align="center">First Name </div></th>
<th width="98"> <div align="center">Last Name </div></th>
<th width="198"> <div align="center">Email </div></th>
<th width="97"> <div align="center">City </div></th>
<th width="97"> <div align="center">State </div></th>
<th width="59"> <div align="center">Country </div></th>
<th width="59"> <div align="center">View </div></th>
<tr>
<?php
if (isset($_GET['Students']))
{
while ($row=mysql_fetch_array($query))
{
echo $row['email'];
echo $row['address_city'];
echo $row['address_state'];
echo $row['address_country'];
}
}
?>
你有这个自己写的或只是复制粘贴它的花括号结束了吗? – Ikari
@Saitama为什么这个奇怪的问题?为了回答你的问题 - 我正在观看youtube上的视频并遵循代码的大部分内容。这里没有粘贴复制品。一切都是自学的 –
你的代码布局很糟糕,请尝试使用一些标签?您还没有关闭您的
回答
您在选择写一个错误的代码。使用下面的一个。
,你也缺少while循环
来源
2016-03-27 07:02:18 Ajay
它仍然不起作用 –
你能看到选择框吗? – Ajay
是的,这解决了它 –
你缺少}
来源
2016-03-27 07:03:23 Thiyagesan
是的,这解决了它。 –
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