0
我试图使用三个部门显示MySQL结果（有和没有搜索）。 Div 1具有用于选择查看结果的单选按钮。 Div 2有一个文本说明所有成员都显示，名称的文本搜索框和类型的下拉菜单。搜索结果显示在Div 3中。无法在不同的div中显示来自搜索的Mysql搜索结果
选择所有成员后，文本“显示所有成员”显示在Div 2中，正确的数据显示在Div 3中（正常工作）。当选择名称单选按钮正确的文本搜索框显示Div 2，但我收到以下错误Div 3：
警告：mysqli_query（）：无法获取/ home /上线/member_dir_test2.php 171
警告：mysqli_error（）期望的是1个参数，0在/home/desgar20/elrenochamber.com/member_dir_test2.php给定线173 无法访问数据库：
如果我在搜索框中输入文字并提交Div 2和Div 3去空白
我是使用JavaScript来显示/隐藏搜索的分部和php。
我已经搜索了答案，但还没有找到任何解决此问题的方法。我不知道我错过了什么。需要帮助的建议。
<style type="text/css">
.box {
display: none;
}
</style>
<script type="text/javascript">
//Show or application part based on selection
$(document).ready(function(){
$('input[type="radio"]').click(function(){
if($(this).attr("value")=="all"){
$(".box").not(".all").hide();
$(".all").show();
$(".all_listing").show();
}
if($(this).attr("value")=="name"){
$(".box").not(".name").hide();
$(".name").show();
$(".all_name").show();
}
if($(this).attr("value")=="type"){
$(".box").not(".type").hide();
$(".type").show();
$(".all_type").show();
}
});
});
</script>
</head>
<body>
<?php
require_once 'php/dbconnect.php'; // Connect to database
$connection = db_connect();
?>
<div id="container">
<div id="service">
<div id="web">
<img width="150px" src="images/search1.png" />
<h3>Member <strong><span class="green">Directory</span></strong></h3>
<strong>View Members By:</strong><br /><br />
<div id="sortOptions">
<label><input type="radio" name="sortRadio" value="all"> All Members</label><br />
<label><input type="radio" name="sortRadio" value="name"> Name</label><br />
<label><input type="radio" name="sortRadio" value="type"> Type</label>
</div><!-- sortOptions -->
</div><!-- end web -->
<div id="vector">
<div class="all box">
<h3>Display <strong><span class="green">All Members</span></strong></h3>
<p>All Members Displayed</p>
</div><!-- all box -->
<div class="name box">
<h3>Display by <strong><span class="green">Member Name</span></strong></h3>
<p><form name="namesearch" method="post" action="<?php echo htmlentities($_SERVER['PHP_SELF']);?>">
Name: <input type="text" name="find">
<input type="submit" name="search" value="Search Names">
</form></p>
</div><!-- name box -->
<div class="type box">
<h3>Display by <strong><span class="green">Member Type</span></strong></h3>
<p><form action="<?php echo htmlentities($_SERVER['PHP_SELF']);?>" method="post">
Type: <select name="type" id="type">
<option value="">-- Select A Type --</option>
<?php
$query = "SELECT * FROM select_type"; //create type drop-down menu
$result = mysqli_query($connection, $query);
while ($line = mysqli_fetch_array($result)) {
echo "<option value='". $line['type'] ."'>". $line['type']."</option>";
}
?>
</select>
<input type="submit" name="searchType" value="Search Types">
</form>
</div><!-- type box -->
</div><!-- end vector -->
</div><!-- end service-->
<div id="media" class="group">
<div class="all_listing box">
<p>Directory Listing</p>
<!-- Start Directory Listing -->
<?php
$sql = "SELECT * FROM members ORDER BY name"; // Database query and results
$result = mysqli_query($connection, $sql);
while($row = mysqli_fetch_array($result)) {
// Check record for website
if ($row['web']!== "") {
echo "<a target=blank href=". $row['web'] . ">" .$row["name"] . "</a><br> " .
"". $row["type"] . "<br>" .
"Address: " . $row["physicaladdress"] . "<br>" .
"Phone: " . $row["phone"] . "<br>" . "<hr>";
}
else {
echo $row["name"]. "<br> " .
"". $row["type"] . "<br>" .
"Address: " . $row["physicaladdress"] . "<br>" .
"Phone: " . $row["phone"] . "<br>" . "<hr>";
}
}
$connection->close();
?>
</div><!-- all_listing box -->
<div class="all_name box">
<p>Results based on Name Search</p>
<?php
if (isset($_POST['search'])) { // Has "Select Names button ben pushed
$find = $_POST['find'];
$sql = "SELECT * FROM members WHERE name LIKE '%" . $find . "%' ";
$result = mysqli_query($connection, $sql);
if(! $result) {
die ('Could not access database: ' . mysqli_error());
}
while($row = mysqli_fetch_array($result)) {
// Check record for website
if ($row['web']!== "") {
echo "<a target=blank href=". $row['web'] . ">" .$row["name"] . "</a><br> " .
"". $row["type"] . "<br>" .
"Address: " . $row["physicaladdress"] . "<br>" .
"Phone: " . $row["phone"] . "<br>" . "<hr>";
}
else {
echo $row["name"]. "<br> " .
"". $row["type"] . "<br>" .
"Address: " . $row["physicaladdress"] . "<br>" .
"Phone: " . $row["phone"] . "<br>" . "<hr>";
}
}
}
?>
</div><!-- all_name box -->
</body>
</html>
我修改了代码，作为建议，但仍收到上述相同的错误，提。下面是我有困难的地区：
这是划分（DIV 1）将在2区显示搜索选项：
<div id="web">
<div id="sort_options">
<label><input type="radio" name="sortRadio" value="name"> Name</label><br />
</div><!-- sort_options -->
</div><!-- end web -->
这是2区的搜索功能：
<div class="name box">
<form name="namesearch" method="post" action="<?php echo htmlentities($_SERVER['PHP_SELF']);?>">
Name: <input type="text" name="find">
<input type="submit" name="search" value="Search Names">
</form></p>
</div>
这是PHP代码在DIV 3从DIV 2运行与输入的查询：
<div class="all_name box">
<?php
if (isset($_POST['search']))
{
$find = $_POST['find'];
$sql = "SELECT * FROM members WHERE name LIKE '%" . $find . "%' ";
$result = mysqli_query($connection, $sql);
if ($result == false)
{
echo ("Error description: " . mysqli_error($connection));
}
else
{
while($row = mysqli_fetch_array($result))
{
// Check record for website
if ($row['web']!== "")
{
echo "<a target=blank href=". $row['web'] .">" .$row["name"] . "</a><br> " .
"". $row["type"] . "<br>" .
"Address: " . $row["physicaladdress"] . "<br>" .
"Phone: " . $row["phone"] . "<br>" . "<hr>";
}
else
{
echo $row["name"]. "<br> " .
"". $row["type"] . "<br>" .
"Address: " . $row["physicaladdress"] . "<br>" .
"Phone: " . $row["phone"] . "<br>" . "<hr>";
}
}
}
}
?>
</div>
我在这里做错了什么？
你真的应该把这段代码分成合适的层次...... – blackbird
总是对学习更聪明地工作感兴趣。我会研究适当的图层。非常感谢 – Ron