我目前正在自学 PHP，现在，我正在尝试创建一个简单的图书馆系统。现在，我在使用搜索选项时遇到了一些问题。我相信我的查询和条件语句是正确的，但我仍然无法显示输出。
这是代码:
<html>
<link rel="stylesheet" type="text/css" href="style.css">
<head>Home Page!</head>
<h1>List of Books</h1>
<?php include 'show.php'; ?>
<?php
require 'dbcon.php';
$Terror = $Aerror = $Derror = $Cerror = $matches = "";
if ($_SERVER["REQUEST_METHOD"] == "POST"){
$T = $_POST['title'];
$A = $_POST['Author'];
$D = $_POST['Desc'];
$C = $_POST['Category'];
$X = $_POST['Opt'];
$del_sql = "delete from a where Title = '".$T."' AND Author = '".$A."' ";
"delete from b where Title = '".$T."' AND Description = '".$D."' AND Category = '".$C."' ";
if($X == 'Add'){
$in_sql = "insert a (Title, Author) values ('".$T."', '".$A."'); insert b (Title, Description, Category) values ('".$T."', '".$D."', '".$C."')";
if(empty($T) || empty($A) || empty($D) || empty($C)){
if(empty($A)) $Aerror = "Fill up Author";
if(empty($T)) $Terror = "Fill up Title";
if(empty($D)) $Derror = "Fill up Description";
if(empty($C)) $Cerror = "Fill up Category";
}
else if(mysqli_multi_query($dbcon, $in_sql)){
echo "New recored added";
}
}
else if ($X == 'Del'){
if (mysqli_multi_query($dbcon, $del_sql)){
echo "Record deleted";
}
}
else if ($X == 'Search'){
$sea1_sql = "select count(Title) from a where Title = '".$T."'";
$sea_sql = mysqli_multi_query($dbcon, "select * from a where Title = '".$T."' AND Author = '".$A."'");
"select Description, Category from b where Title = '".$T."' AND Description = '".$D."' AND Category = '".$C."'";
if (mysqli_query($dbcon, $sea1_sql) > 0){
while ($row = mysqli_fetch_array($sea_sql)){
echo "<table class = fix>";
echo "<tr><th>Title</th> <th> ". $row['Title'] . " </th></tr>";
echo "<tr><th>Author </th> <th> ". $row['Author'] . " </th></tr>";
echo "<tr><th>Description </th> <th> ". $row['Description'] . " </th></tr>";
echo "<tr><th>Category </th> <th> ". $row['Category'] . " </th></tr>";
echo "</table><br>";
}
}
else if (mysqli_query($dbcon, $sea1_sql) < 0){
echo "Can't find data!";
}
}
}
?>
<body>
<form method="post" action="<?php echo htmlspecialchars($_SERVER["PHP_SELF"]);?>">
<br>Title: <input type="text" name="title"> <?php echo $Terror; ?><br>
Author: <input type="text" name="Author"><?php echo $Aerror; ?><br>
Description: <input type="text" name="Desc"><?php echo $Derror; ?><br>
Category: <input type="text" name="Category"><?php echo $Cerror; ?><br>
<select name="Opt">
<option disabled selected>N/A</option>
<option value='Add'>Add</option>
<option value='Edit'>Edit</option>
<option value='Del'>Del</option>
<option value='Search'>Search</option>
</select><br>
<input type="submit" name="submit">
</form>
</body>
</html>
最佳答案
这是错误的:
if (mysqli_query($dbcon, $sea1_sql) > 0){
myqli_query() 不返回 count(Title) 的值。您需要获取该行。
$sea1_result = mysqli_query($dbcon, $sea1_sql);
$row = mysqli_fetch_assoc($sea1_result);
if ($row['count(Title)'] > 0) {
您还错误地使用了 mysqli_multi_query()。它不返回您可以从中获取结果的 mysqli_result 对象，它返回一个 bool 值，指示第一个查询是成功还是出错。您需要调用mysqli_use_result()获取每个查询的结果，然后在此调用 mysqli_fetch_array()。
但看起来您根本不应该使用 mysqli_multi_query()。您想要从 a 和 b 中获取相同标题的行，而不是所有 a 行都在所有 b 之前> 行。你应该加入两个表:
$sea_result = mysqli_query("SELECT a.*, b.Description, b.Category
FROM a JOIN b ON a.Title = b.Title
WHERE a.Title = '$T' AND a.Author = 'A'
AND b.Description = '$D' AND b.Category = '$C'");
关于Php mysql 搜索选项，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/30904927/