我想将值存储在变量中然后显示在表格中,但这不会起作用

时间:2016-08-11 18:29:34

标签: php mysql

我想先将结果存储在变量中,以便我可以使用它们,但这不会迭代任何东西。请参阅我尝试先将行的值存储在变量中并在表的td中调用它们,但它没有显示任何内容。

是否存在语法错误或其他问题。什么可能是首先存储在变量中的方法?

 $query_test = "SELECT itemname, categoryname, manufacturername,
                       price, shopname,itemurl, itemimage, typename  
                FROM prices p, 
                     items i, 
                     shops s, 
                     categories c, 
                     manufacturers m, 
                     types t, 
                    modules mo 
                WHERE p.shopid=s.shopid 
                  AND i.categoryid=c.categoryid 
                  AND p.itemid=i.itemid 
                  AND i.ManufacturerId=m.ManufacturerID 
                  AND i.ModuleId= mo.ModuleID 
                  AND i.TypeId=t.TypeID 
                  AND i.categoryid=1004 
                  AND s.shopid=5003";

 $result = mysqli_query($conn, $query_test);
 echo "<table border='1'>";
 echo "<tr><td>"."Name"."</td>";
 echo "<td>"."Category"."</td>";
 echo "<td>"."Manufacturer"."</td>";
 echo "<td>"."Price"."</td>";
 echo "<td>"."Shop"."</td>";
 echo "<td>"."Type"."</td>";
 while ($row = mysqli_fetch_array($result, MYSQLI_BOTH)) {
    //output a row here
    $name = $row['itemname'];
    $category = $row['categoryname'];
    $manufacturer = $row['manufacturername'];
    $price = $row['price']:
    $shop = $row['shopname'];
    $type = $row['typename'];
    echo "<tr><td>".$name."</td>";
    echo "<td>".$category."</td>";
    echo "<td>".$manufacturer."</td>";
    echo "<td>".$price."</td>";
    echo "<td>".$shop."</td>";
    echo "<td>".$type."</td></tr>";
    /*echo "<tr><td>".($row['itemname'])."</td>";
    echo "<td>".($row['categoryname'])."</td>";
    echo "<td>".($row['manufacturername'])."</td>";
    echo "<td>".($row['price'])."</td>";
     echo "<td>".($row['shopname'])."</td>";
    echo "<td>".($row['typename'])."</td></tr>";*/
 }

 echo "</table>";

1 个答案:

答案 0 :(得分:1)

在while循环中执行var_dump($row),以查看$row中存储的值。

另外,考虑到你如何访问它们,你应该这样做:

while( $row = mysqli_fetch_assoc($result))