我正在使用PHP从我的数据库中获取一些记录,它一直返回“没有找到结果”。但是,当我将查询打印到屏幕上并将其复制并粘贴到mysql提示符处时,我得到1行。当然,这是一个怪物查询,但不应该得到相同的结果吗?是什么导致这种情况发生?有什么建议要检查什么?mysql - 在mysql提示符下获取结果,但不通过脚本?
不知道,如果它是有帮助的,但这里的查询:
$q = db_query("SELECT node.nid AS nid, gallery.field_attach_gallery_value AS gallery,
node.type AS type, node.vid AS vid, ce.field_brochure_link_url AS ce_brochure_url,
ce.field_brochure_link_title AS ce_brochure_title,
ce.field_brochure_link_attributes AS ce_brochure_attributes,
location.field_location_value AS location_field_location_value,
ce.field_ongoing_value AS ce_field_ongoing_value,
ce.field_poster_link_url AS ce_poster_url,
ce.field_poster_link_title AS ce_poster_title,
ce.field_poster_link_attributes AS ce_poster_attributes,
ce.field_press_release_link_url AS ce_press_release_url,
ce.field_press_release_link_title AS ce_press_release_title,
ce.field_press_release_link_attributes AS ce_press_release_attributes,
(DATE_FORMAT(STR_TO_DATE(ce.field_start_date_value,'%Y-%m-%dT%T'),'%M %d, %Y %h:%i %p')) AS start,
(DATE_FORMAT(STR_TO_DATE(ce.field_start_date_value2,'%Y-%m-%dT%T'),'%M %d, %Y %h:%i %p')) AS end,
ce.field_web_resources_url AS ce_web_resources_url,
ce.field_web_resources_title AS ce_web_resources_title,
ce.field_web_resources_attributes AS ce_web_resources_attributes,
node_revisions.body AS body,
node_revisions.format AS node_revisions_format,
node.title AS title
FROM node node
LEFT JOIN content_field_attach_gallery gallery ON node.vid = gallery.vid
LEFT JOIN content_type_exhibitions_and_programs ce ON node.vid = ce.vid
LEFT JOIN content_field_location location ON node.vid = location.vid
LEFT JOIN node_revisions node_revisions ON node.vid = node_revisions.vid
WHERE (node.status <> 0)
AND (node.type in ('exhibitions_and_programs'))
AND '2010-01-19' BETWEEN (DATE_FORMAT(STR_TO_DATE(ce.field_start_date_value, '%Y-%m-%dT%T'), '%Y-%m-%d'))
AND (DATE_FORMAT(STR_TO_DATE(ce.field_start_date_value2, '%Y-%m-%dT%T'), '%Y-%m-%d'))
ORDER BY (DATE_FORMAT(STR_TO_DATE(ce.field_start_date_value, '%Y-%m-%dT%T'), '%Y-%m-%d')) DESC");
$num = FALSE;
while($r = db_fetch_array($q)) {
$num = TRUE;
$line = 'ok, found something!';
}
if($num == TRUE) {
print $line;
} else {
print 'No records found';
}
var_dump只是给了我:“警告:mysql_fetch_array():提供的参数不是在/usr/local/www/s/drupal-6.14/includes/database.mysql.inc在线160上的有效MySQL结果资源。”, ,我已经知道它没有工作,但是,我不明白为什么它能正常工作,如果我只是在提示符下运行完全相同的查询? – phpN00b 2010-01-19 17:25:54
佩卡的编辑答案是正确的。您需要在查询中手动转义'%d'(而不是%b),因为Drupal可能试图将这些解释为字符串参数。 – loginx 2010-01-19 17:34:09
干杯@loginx,纠正了我的错误。 – 2010-01-19 18:43:10