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//$type has value of "Hello+World"
$type = $_POST['series'];

$sql = "select max(id) from TABLE_NAME where type = " . $type;

$result = sybase_query ($sql, $db_ro_conn) or die(db_error("query failed $sql"));
$row = sybase_fetch_row($result)

我在 $sql 行中收到错误““=”附近的语法不正确。y15,过程 N/A。

发生这种情况的可能原因是什么?不知何故,它不起作用。将不胜感激任何帮助,谢谢!

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3 回答 3

1

像这样在 $type 周围加上引号:

$type = $_POST['series'];

$sql = "select max(id) from TABLE_NAME where type = '" . $type. "'";

$result = sybase_query ($sql, $db_ro_conn) or die(db_error("query failed $sql"));
$row = sybase_fetch_row($result)
于 2012-06-01T11:29:45.180 回答
0

让我从这个开始。始终在查询中转义 POST/GET 值!

该错误可能是由于 $type 是字符串而不是引号引起的。尝试将 $sql 更改为

$sql = "select max(id) from TABLE_NAME where type = '" . $type."'";
于 2012-06-01T11:30:08.087 回答
0

您没有引用该值:

$type = str_replace("'", "''", $_POST['series']);
$sql = "select max(id) from TABLE_NAME where type = '" . $type . "'";
于 2012-06-01T11:30:57.590 回答