為什么PHP無法識別此查詢？
[英]Why doesn't php recognize this query?
問題描述
我試圖在php中運行一個簡單的mysql查詢，但是沒有運行。 它說B.ISBN等是未定義的索引。 這是為什么 ？ php無法識別mysql的AS（某些東西）嗎？ 我已經編輯過要添加文件的完整代碼，對於任何誤解，我們深表歉意
<?php include 'dbconfig.php'; $Title = $_GET['Title'];?>
<?php header('Content-Type: text/html; charset=utf-8');?>
<!DOCTYPE HTML PUCLIC "-//W3C//DTDHTML 4.0 Transitional//EN">
<HTML>
<HEAD>
<link rel="stylesheet" type="text/css" href="mytable.css">
<link rel="stylesheet" type="text/css" href="styles.css">
</HEAD>
<TABLE class="minimalistBlack">
<thead>
<tr>
<th> ISBN </th>
<th> Title </th>
<th> Number of Pages </th>
<th> Publisher Name </th>
<th> Copy Number </th>
<th> Shelf </th>
</tr>
</thead>
<?php
$conn= mysqli_connect("localhost","root","","library");
mysqli_set_charset($conn, "utf8");
if ($conn -> connect_error){
die("Conenction failed:". $conn->connect_error);
}
$sql="SELECT B.ISBN,B.Title,B.numpages,B.pubName,C.copyNr,C.shelf
FROM book AS B
INNER JOIN copies as C
ON B.ISBN=C.ISBN";
$result = $conn->query($sql);
if ($result->num_rows>0){
while($row= $result->fetch_assoc()){
echo "<tr>";
echo "<td>".$row['B.ISBN']."</td>";
echo "<td>".$row['B.Title']."</td>";
echo "<td>".$row['B.numpages']."</td>";
echo "<td>".$row['B.pubName']."</td>";
echo "<td>".$row['C.copyNr']."</td>";
echo "<td>".$row['C.shelf']."</td>";
}
echo "</TABLE>";
}
else { echo "0 result"; }
$conn->close();
?>
</TABLE>
</HTML>
4 個解決方案
解決方案1
1 已采納 2019-05-25 17:41:27
使用JOIN ：
SELECT B.ISBN,B.Title,B.numpages,B.pubName,C.copyNr,C.shelf
FROM book AS B
INNER JOIN copies as C
ON B.ISBN=C.ISBN
WHERE B.Title='$Title'";
解決方案2
0 2019-05-25 17:44:48
您可以使用此命名法命名表。
SELECT B.ISBN,B.Title,B.numpages,B.pubName FROM book B JOIN copies C WHERE B.ISBN=C.ISBN AND B.Title = '+$title;
解決方案3
0 2019-05-25 17:44:49
請嘗試以下查詢：
SELECT B.ISBN,
B.Title,
B.numpages,
B.pubName,
C.copyNr,
C.shelf
FROM book as B
LEFT JOIN copies AS C ON B.ISBN=C.ISBN
WHERE B.Title = '”.$Title.”’”
解決方案4
0 2019-05-25 17:56:54
$sql="SELECT B.ISBN,B.Title,B.numpages,B.pubName,C.copyNr,C.shelf
FROM book AS B
INNER JOIN copies as C
ON B.ISBN=C.ISBN
WHERE B.Title = '".$Title."'";
為什么我的PHP無法識別我的文件？ Isset
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