我只有两个数据库，它们都运行在同一个选择脚本中。该脚本适用于一个，但另一个只显示数据库的前三行，并重复显示它们。
由于我是PHP的新手，我无法想象为什么它适用于一个数据库而不适用于另一个数据库。我检查了所有拼写，然后我回答了查询脚本以确定。更糟糕的是，浏览器并没有给我一个错误作为线索。它只是显示错误的信息。
请帮忙。 。谢谢你！
index.html（这是形式......我不认为这个问题在这里......但我包括它以防万一...我认为问题必须在product_list.php中删除）< / p>
<!DOCTYPE html>
<html>
<head><title>Databases</title></head>
<body>
<h1>Music Store Database</h1>
<form method='POST' action='display.php'
<label>Select a table:</label>
<select name="tableName">
<option value="products">Products</option>
<option value="categories">Categories</option>
</select>
<p>To ADD: Enter the field values(s) below for the record(s) you want to add. NOTE: Adding a record requires all appropriate field for the selected table.<p>
<p>To DELETE: Enter the value for the field you are going to use to identify the desired record(s). Then select that field from "delete record(s)" section below.</$
<label>ProductID</label>
<input type="text" name="productIDtx" value=""/></br>
<label>CategoryID</label>
<input type="text" name="categoryIDtx" value=""/></br>
<label>Product Code</label>
<input type="text" name="productCodetx" value=""/></br>
<label>Product Name</label>
<input type="text" name="productNametx" value=""/></br>
<label>List Price</label>
<input type="text" name="listPricetx" value=""/></br>
<label>Category Name</label>
<input type="text" name="categoryNametx" value=""/></br>
<p>Delete Record(s): Select desired field below. Don't forget to complete the information for the record in question above:</p>
<input type="radio" name="remove" value="productID"/>
<label>ProductID</label><br />
<input type="radio" name="remove" value="categoryID"/>
<label>CategoryID</label><br />
<input type="radio" name="remove" value="productCode"/>
<label>Product Code</label><br />
<input type="radio" name="remove" value="productName"/>
<label>Product Name</label><br />
<input type="radio" name="remove" value="listPrice"/>
<label>List Price</label><br />
<input type="radio" name="remove" value="categoryName"/>
<label>Category Name</label><br />
<p>To Retrieve Record(s): Select field(s) below you want to see from the list below.</p>
<input type="checkbox" name="productIDcb"/>
<label>ProductID</label><br />
<input type="checkbox" name="categoryIDcb"/>
<label>CategoryID</label><br />
<input type="checkbox" name="productCodecb"/>
<label>Product Code</label><br />
<input type="checkbox" name="productNamecb"/>
<label>Product Name</label><br />
<input type="checkbox" name="listPricecb"/>
<label>List Price</label><br />
<input type="checkbox" name="categoryNamecb"/>
<label>Category Name</label><br />
<p>Select the appropriate action based on your selection from above:</p>
<input type="radio" name="operation" value="retrieve"/>
<label>Retrieve Information</label><br />
<input type="radio" name="operation" value="addition"/>
<label>Add Information</label><br />
<input type="radio" name="operation" value="delete"/>
<label>Delete Information</label><br />
<p><input type="submit" value="Submit Request"/></p>
</form>
</body>
</html>
display.php（这会从我的表单中捕获信息并调用函数...仍然包括以防万一，但我想问题必须是下一个文件。）
<?php
require('database.php');
$productIDtx = $_POST['productIDtx'];
$categoryIDtx = $_POST['categoryIDtx'];
$productCodetx = $_POST['productCodetx'];
$productNametx = $_POST['productNametx'];
$listPricetx = $_POST['listPricetx'];
$categoryNametx = $_POST['categoryNametx'];
if(isset($_POST['tableName']))
{
$table = $_POST['tableName'];
}
else
{
echo("Must select a table.<br>");
}//endif
if(isset($_POST['operation']))
{
$operation = $_POST['operation'];
}
else
{
echo("Must select an action.<br>");
exit();
}//endif
if(isset($_POST['remove']))
{
$remove = $_POST['remove'];
}
else
{
$remove = "";
}//endif
if(isset($_POST['productIDcb']))
{
$productIDcb = $_POST['productIDcb'];
}
else
{
$productIDcb = "";
}//endif
if(isset($_POST['categoryIDcb']))
{
$categoryIDcb = $_POST['categoryIDcb'];
}
else
{
$categoryIDcb = "";
}//endif
if(isset($_POST['productCodecb']))
{
$productCodecb = $_POST['productCodecb'];
}
else
{
$productCodecb = "";
}//endif
if(isset($_POST['productNamecb']))
{
$productNamecb = $_POST['productNamecb'];
}
else
{
$productNamecb = "";
}//endif
if(isset($_POST['listPricecb']))
{
$listPricecb = $_POST['listPricecb'];
}
else
{
$listPricecb = "";
}//endif
if(isset($_POST['categoryNamecb']))
{
$categoryNamecb = $_POST['categoryNamecb'];
}
else
{
$categoryNamecb = "";
}//endif
if($operation == 'retrieve')
{
include_once('product_list.php');
show_products($productIDcb, $categoryIDcb, $productCodecb, $productNamecb, $listPricecb, $categoryNamecb);
}
elseif($operation == 'addition')
{
include_once('addprod.php');
add($table, $productIDtx, $categoryIDtx, $productCodetx, $productNametx, $listPricetx, $categoryNametx);
}
elseif($operation == 'delete')
{
include_once('deleteprod.php');
delete($table, $remove, $productIDtx, $categoryIDtx, $productCodetx, $productNametx, $listPricetx, $categoryNametx);
}
else
{
echo('<p>Select an action: Retrieve, Add, or Delete. </p>'); //this code shouldn't ever happen because $operation is tested above but I put it in here in case an errors at this point in the code.
exit();
}//endif
?>
product_list.php（我认为问题必须是$ rSET或之后的地方，因为$ theQuery的回显显示正确。但这适用于其他数据库。所以我不知所措。）
<?php
include('database.php');
function show_products($productIDcb, $categoryIDcb, $productCodecb, $productNamecb, $listPricecb, $categoryNamecb)
{
global $db;
$theQuery = 'select ';
$list = "";
if($productIDcb == "")
{
$theQuery == $theQuery;
}
else
{
$theQuery .= 'p.productID, ';
}//endif
if($categoryIDcb == "")
{
$theQuery == $theQuery;
}//endif
else
{
$theQuery .= 'c.categoryID, ';
}//endif
if($productCodecb == "")
{
$theQuery == $theQuery;
}
else
{
$theQuery .= 'p.productCode, ';
}//endif
if($productNamecb == "")
{
$theQuery == $theQuery;
}
else
{
$theQuery .= 'p.productName, ';
}//endif
if($listPricecb == "")
{
$theQuery == $theQuery;
}
else
{
$theQuery .= 'p.listPrice, ';
}//endif
if($categoryNamecb == "")
{
$theQuery == $theQuery;
}
else
{
$theQuery .= 'c.categoryName, ';
}//endif
$theQuery .=" '' from (categories c, products p) where (c.categoryID = p.categoryID);";
echo($theQuery);
echo('<br>');
//***I THINK THE ISSUE MUST BE SOMEWHERE AFTER THIS***
$rSet = $db -> query($theQuery);
foreach($rSet AS $results)
{
$list .=' '.$results[0];
if(isset($results[1]))
{
$list .=' '.$results[1];
}
if(isset($results[2]))
{
$list .=' '.$results[2];
}
if(isset($results[3]))
{
$list .=' '.$results[3];
}
if(isset($products[4]))
{
$list .=' '.$results[4];
}
$list .="<br>";
}//end foreach
echo($list);
echo('<br>');
echo('<a href="index.html">Music Store Database</a>');
}//end function
?>
产品数据库查询（这个工作）
选择p.productID，c.categoryID，p.productCode，p.productName，p.listPrice，''from（categories c，products p）where（c.categoryID = p.categoryID）;
<1> 1 strat Fender Stratocaster2 1 les_paul Gibson Les Paul
3 1 sg Gibson SG
4 1 fg700s雅马哈FG700S
5 p Washburn Washburn D10S6 1 rodriguez Rodriguez Caballero 11
7 2精密挡泥板精度
8 2 hofner Hofner Icon
9 3路德维希路德维希五件套鼓镲套装
10 3 tama Tama带镲片的5件套鼓
类别数据库查询（查询看起来正确但数据错误）
选择c.categoryID，c.categoryName，''from（categories c，products p）where（c.categoryID = p.categoryID）;
1吉他
1吉他
1吉他
1吉他
1吉他
1吉他
2 Basses
2 Basses
3鼓
3鼓
它应列出以下内容
1吉他
2 Basses
3鼓
11测试
15测试
20测试
33 test33
40测试
（注意：测试来自我测试添加功能的时候 - 如果你想知道的话）。
答案 0 :(得分：0)
您的查询将导致笛卡尔积，它会返回重复的结果。请尝试使用INNER JOIN或CROSS JOIN。
SELECT categoryID, categoryName
FROM categories c
CROSS JOIN products p
WHERE (c.categoryID = p.categoryID)
答案 1 :(得分：0)
感谢上面评论过的所有人，让我明白这一点。正如andrewsi指出的那样，SELECT DISTINCT代码确实修复了重复的问题，但由于（c.categoryID = p.categoryID），整个类别表都不会显示。所以我添加了if if语句来处理这两种情况，这也消除了对DISTINCT的需求。这是修复它的代码。
if($theQuery == 'select c.categoryID, c.categoryName, ')
{
$theQuery = "";
$theQuery = "select * from categories;";
}
else
{
$theQuery .=" '' from (categories c, products p) WHERE (c.categoryID = p.categoryID);";
}