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Php mysql drupal 6查询问题_Php_Mysql_Drupal_Drupal 6 - Fatal编程技术网

Php mysql drupal 6查询问题

Php mysql drupal 6查询问题,php,mysql,drupal,drupal-6,Php,Mysql,Drupal,Drupal 6,我试图查询具有最低数值的条目,然后将数据库中的数值更新为值+1,如下所示: $ras = db_query("SELECT name,number FROM {people} ORDER BY number DESC LIMIT 1"); $raw = db_fetch_array($ras); $com_name = $raw['name']; $count = $raw['number']+1; db_query("UPDATE {people} (number) SET ('%d') W

我试图查询具有最低数值的条目,然后将数据库中的数值更新为值+1,如下所示:

$ras = db_query("SELECT name,number FROM {people} ORDER BY number DESC LIMIT 1"); 
$raw = db_fetch_array($ras);
$com_name = $raw['name'];
$count = $raw['number']+1;
db_query("UPDATE {people} (number) SET ('%d') WHERE name='%s'", $count, $com_name);

我得到一个错误:您的SQL语法有一个错误;检查与您的MySQL服务器版本对应的手册,以获取正确的语法。这里怎么了?

您的
SET
子句错误

试试这个:

db_query("UPDATE {people} SET number = '%d' WHERE name='%s'", $count, $com_name);
或者,如果您的
number
列是
INT
类型:

db_query("UPDATE {people} SET number = %d WHERE name='%s'", $count, $com_name);

注意删除的引号?

您的
SET
子句是错误的

试试这个:

db_query("UPDATE {people} SET number = '%d' WHERE name='%s'", $count, $com_name);
或者,如果您的
number
列是
INT
类型:

db_query("UPDATE {people} SET number = %d WHERE name='%s'", $count, $com_name);

注意删除的引号?

这很有效,谢谢!在我接受答案之前几分钟。是否将%d置于引号中有关系吗?@ingrid否。没关系。这很有效,谢谢!在我接受答案之前几分钟。是否将%d置于引号中有关系吗?@ingrid否。没关系。