Question

我有一个包含MyPrice，MyStock等列的数据库。每个产品有几行 - 苹果，桃子等。

我正在尝试使用以下代码提取有关多个项目的价格/库存状态的数据，但我得到的错误如下：

Notice: Undefined variable: sql on line 11

Warning: mysqli::query(): Empty query on line 11

Notice: Undefined variable: prices on line 45

Notice: Trying to get property of non-object on line 45

$dbhost =   'zzz';
$dbuser =   'zzz'; 
$dbpwd  =   'zzz'; 
$dbname =   'zzz';

$conn = new mysqli( $dbhost, $dbuser, $dbpwd, $dbname );                
$res = $conn->query( $sql );

$item1 = "apple";
$item2 = "peach";

$items=array( $item1, $item2 );
$sql='select * from products where `Name` in ("'.implode('","',$items).'");'; //line 11

if( $res ){

    $i=0;/* counter for dynamic variables */
    $prices=new stdClass;

    while( $rs=$res->fetch_object() ){
        $i++;

        /* Populate an object with details for the product */
        $prices->{ strtolower( $rs->name ) }=(object)array( 'price'=>$rs->MyPrice, 'description'=>$rs->MyDescription, 'stock'=>$rs->MyStock );

    }
}   
?>  

echo 'Apple:'.$prices->apple->price . '<br />'; //line 45
echo 'Peach:'.$prices->peach->stock . '<br />';

导致问题的原因是什么？

Answer 1

我已根据您获得的错误调试了您的代码。无论如何，信息应该引导你自己解决问题，特别是如果它们清晰明显的话。如果您总是让其他人调试您的代码，那么您将无法学到任何东西。

db.collections.aggregate([ 
    { "$project": { "count": { "$size": "$Documents" } } } 
] )

Answer 2

移动$ res = $ conn-＆gt;查询（$ sql）; 第12行

同时删除结尾？＆GT; 并在文件末尾放置

Answer 3

这是你的第一个罪犯：

$sql='select * from products where `Name` in ("'.implode('","',$items).'");';

在尝试将其连接到sql语句之前，你应该破坏$ items。

类似的东西：

$item_text = implode('","',$items);
$sql='select * from products where `Name` in ("'.$item_text.'");';

老实说，这段代码有点混乱。您应该使用PDO或mysql而不是两者都清理它。

Answer 4

这是一个开始

$dbhost =   'zzz';
$dbuser =   'zzz'; 
$dbpwd  =   'zzz'; 
$dbname =   'zzz';

$conn = new mysqli( $dbhost, $dbuser, $dbpwd, $dbname );                

$item1 = "apple";
$item2 = "peach";

$items=array( $item1, $item2 );
$item_text = implode('","',$items);
$sql='select * from products where `Name` in ("'.$item_text.'");'; //line 11

$res = $conn->query( $sql );

if( $res ){

    $i=0;/* counter for dynamic variables */
    $prices=new stdClass;

    while( $rs=$res->fetch_object() ){
        $i++;

        /* Populate an object with details for the product */
        $prices->{ strtolower( $rs->name ) }=(object)array( 'price'=>$rs->MyPrice, 'description'=>$rs->MyDescription, 'stock'=>$rs->MyStock );

    }
}   


echo 'Apple:'.$prices->apple->price . '<br />'; //line 45
echo 'Peach:'.$prices->peach->stock . '<br />';

?>

MySQL未定义的变量错误

4 个答案: