如何从mysql中的列中选择多个记录(行)？

Question

我想从这些id中显示四(4)项的名称:我可以这样做吗？

SELECT item_name from items WHERE item_id IN ('001', '012', '103', '500')

要么

SELECT item_name from items WHERE item_id = '001' or item_id = '012' or item_id = '103' or item_id = '500'

响应所有答案

好吧,大多数答案说它有效,但它确实不起作用.这是我的代码:

$query = "SELECT `item_name` from items WHERE item_id IN('s001','a012','t103','p500')";

$result = mysql_query($query, $conn) or die (mysql_error());
$fetch =  mysql_fetch_assoc($result) or die (mysql_error());
$itemsCollected = $fetch['item_name'];
echo $itemsCollected;

这item_id是字母数字.

Answer 1

您可以执行任一操作,但对于任何大型查询,IN查询都可以更有效地实现此目的.我很久以前做了一些简单的测试,发现使用IN构造的速度要快10倍.如果你问的是语法是否正确,那么它看起来很好,除了缺少分号以完成语句.

编辑:看起来你问的实际问题是"为什么这些查询只返回一个值".那么,看看你发布的示例代码,问题出在这里:

$fetch = mysql_fetch_assoc($result) or die (mysql_error());
$itemsCollected = $fetch['item_name'];
echo $itemsCollected;

Pax指出,你需要循环并迭代,直到没有更多的结果被提取.请参阅mysql_fetch_assoc的PHP手册页:

$sql = "SELECT item_name from items WHERE item_id IN('s001','a012')";
$result = mysql_query($sql);

if (!$result) {
    echo "Could not successfully run query ($sql) from DB: " . mysql_error();
    exit;
}

if (mysql_num_rows($result) == 0) {
    echo "No rows found, nothing to print so am exiting";
    exit;
}

// While a row of data exists, put that row in $row as an associative array
// Note: If you're expecting just one row, no need to use a loop
// Note: If you put extract($row); inside the following loop, you'll
//       then create $userid, $fullname, and $userstatus
while ($row = mysql_fetch_assoc($result)) {
    echo $row["userid"];
    echo $row["fullname"];
    echo $row["userstatus"];
}

mysql_free_result($result);