编辑：在添加第二个select子句时，对extra“或某种语法有问题。
所以我在使用mysql冒险家样例数据库进行php项目。当前，此页面接收一个员工标题，您可以使用包含所有可能标题的下拉框从中选择该标题以筛选员工。然而，我想添加第二个框与性别，以便有人可以通过过滤标题和性别，并获得所有可能的员工。
我试过一些方法，但通常都会导致页面崩溃。
这里是第一个文件：search.php
<?php
// Connect to the database
if (!include('connect.php')) {
die('error finding connect file');
}
$dbh = ConnectDB();
?>
<html>
<head>
<link href='http://fonts.googleapis.com/css?family=Cuprum'
rel='stylesheet' type='text/css'>
<link href="https://fonts.googleapis.com/css?family=Amaranth"
rel="stylesheet">
<link href="https://fonts.googleapis.com/css?family=Roboto+Mono"
rel="stylesheet">
<link href="main.css" rel="stylesheet">
<title>Table Results</title>
</head>
<body>
<div class="content">
<h1>Search Adventureworks Employees</h1>
<p>Select two parameters and search information from the
AdventureWorks Database.</p>
<div class="form">
<form action="listTable.php" method="get">
<?php
// Get the full list of titles from Employee Table
$sql = "SELECT Distinct title FROM adventureworks.employee";
$stmt = $dbh->prepare($sql);
$stmt->execute();
// Prep drop down control
echo "<label for='title'>Select title: </label>\n";
echo "<select id='title' name='title'>\n";
// Put titles in the options
foreach ($stmt->fetchAll() as $titles) {
echo "<option value='" . $titles['title'] . "'>" .
$titles['title'] . "</option>\n";
}
// End dropdown
echo "</select>";
// Get the full list of genders from employee table
$sql3 = "Select Distinct gender FROM adventureworks.employee";
$stmt = $dbh->prepare($sql3);
$stmt->execute();
//Prep dropdown
echo "<label for ='gender'>Select Gender: </label>\n";
echo "<select id='gender' name = 'gender'>\n";
// Put genders in the options
foreach($stmt->fetchAll() as $genders) {
echo "<option value ='" . $genders['gender'] . "'> .
$genders['gender'] . "</option>\n";
}
//end dropdown and submit
echo "</select> <input type='submit'
value='Submit'>\n</form>\n</div>";
?>
</div>
</body>
</html>下面是第二个php文件：listtable.php
<?php
// Connect to the database
if (!include('connect.php')) {
die('error finding connect file');
}
$dbh = ConnectDB();
?>
<html>
<head><link href='http://fonts.googleapis.com/css?family=Cuprum'
rel='stylesheet' type='text/css'>
<link href="https://fonts.googleapis.com/css?family=Amaranth"
rel="stylesheet">
<link href="https://fonts.googleapis.com/css?family=Roboto+Mono"
rel="stylesheet">
<link href="main.css" rel="stylesheet">
<title>Table Results</title>
</head>
<body>
<div class='content'>
<?php
// Get table name from querystring
if (!isset($_GET["title"])) {
echo "No Title Selected";
}
else {
$title=$_GET["title"];
echo "<h1>Listing of Employees</h1>\n";
$sql1 = "SELECT column_name FROM information_schema.columns ";
$sql1 .= "WHERE table_name = 'employee'";
$stmt = $dbh->prepare($sql1);
$stmt->execute();
$cols = $stmt->rowCount();
// Prep table
$tableHTML = "<table>\n<thead>\n<tr>\n";
// Table headings (column names)
foreach ($stmt->fetchAll() as $columns) {
$tableHTML .= "<th>" . $columns['column_name'] . "</th>\n";
}
// Prep table body
$tableHTML .= "</tr>\n</thead>\n<tbody>\n";
// Table body (column values)
$sql2 = "SELECT * FROM adventureworks.employee e";
$sql2 .= " Where title like '$title'";
$stmt = $dbh->prepare($sql2);
$stmt->execute();
echo $stmt->rowCount() . " rows retrieved<br/><br />\n";
foreach ($stmt->fetchAll() as $rows ) {
$tableHTML .= "<tr>\n";
for ($i = 0; $i < $cols; $i++) {
$tableHTML .= "<td>" . $rows[$i] . "</td>\n";
}
$tableHTML .= "</tr>\n";
}
// End table
$tableHTML .= "</tbody>\n</table>\n";
echo $tableHTML;
echo "<div class='code'><br />" . $sql1 . "<p>" . $sql2 . "</div>\n";
}
?>
</div>
</body>
</html>我知道为了得到我想要的结果，我必须在第二个文件中更改什么，但是我在第一个文件中填充第二个下拉列表时遇到了问题。任何帮助都将不胜感激。谢谢。
1条答案
按热度按时间nukf8bse1#
您在search.php中错过了一个双引号