当我在 PHP 中运行此代码时,结果为空,但是当我在 mysql 终端或 phpmyadmin 中运行它时,我得到了我想要的。
PHP
if ($_GET["action"] == "list") {
//Get records from database
$mainQuery = mysql_query("
SET SQL_BIG_SELECTS=1;
SELECT
ci.id AS item_id,
ar.title, ar.introtext,
flo.value AS logo, fph.value AS phone, fad.value AS address, fur.value AS url, fse.value AS services, fma.value AS map,
ar.id AS joomla_id, ci.hidden_id, ci.type
FROM kd9fb_content ar
RIGHT JOIN calc_settings cs ON ar.catid = cs.joomla_cat
LEFT JOIN kd9fb_fieldsattach_values flo ON flo.articleid = ar.id AND flo.fieldsid = 1
LEFT JOIN kd9fb_fieldsattach_values fph ON fph.articleid = ar.id AND fph.fieldsid = 2
LEFT JOIN kd9fb_fieldsattach_values fad ON fad.articleid = ar.id AND fad.fieldsid = 3
LEFT JOIN kd9fb_fieldsattach_values fur ON fur.articleid = ar.id AND fur.fieldsid = 4
LEFT JOIN kd9fb_fieldsattach_values fse ON fse.articleid = ar.id AND fse.fieldsid = 5
LEFT JOIN kd9fb_fieldsattach_values fma ON fma.articleid = ar.id AND fma.fieldsid = 6
LEFT JOIN calc_item ci ON ci.joomla_id = ar.id
ORDER BY ci.id DESC;
", $con);
while ($row = mysql_fetch_array($mainQuery)) {
$rows[] = $row;
}
//Return result to jTable
$jTableResult = array();
$jTableResult['Result'] = "OK";
$jTableResult['Records'] = $rows;
print json_encode($jTableResult);
}
这将返回 NULL:
$row = mysql_fetch_array($mainQuery)
MySQL基础没问题,代码通过这个连接到基础:
$con = mysql_connect($host, $user, $password) or die("DB login failed!");
mysql_select_db($db, $con) or die("select failed");
mysql_query("SET NAMES utf8");
类似的代码,但其他查询工作正常,我对其进行了测试,显然是请求中的情况。顺便说一句,我认为它是完全非最佳方式构建的,但我不擅长 SQL 和 PHP。
那么问题出在哪里,我哪里出错了?