当我在PHP中运行此代码时,结果为null,但是当我在mysql终端或phpmyadmin中运行它时,我得到了我想要的内容。
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);
}
返回NULL:
$row = mysql_fetch_array($mainQuery)
MySQL基础没问题,代码连接到base:
$con = mysql_connect($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,我测试了它,显然是请求中的情况。顺便说一下,我认为它是完全非最佳的方式,但我不擅长SQL和PHP。
那么问题出在哪里以及我哪里出错了?
答案 0 :(得分:4)
您的查询实际上有两个查询。
"SET SQL_BIG_SELECTS=1;
SELECT
ci.id AS item_id,
ar.title, ar.introtext, ....."
删除查询的第一行,它应该没问题。这是因为mysql_query()方法仅支持一个查询。但我认为终端IIRC已取消了这一限制。
如果您需要第一个语句,只需执行两个查询方法:
mysql_query("SET SQL_BIG_SELECTS=1");
$query = "..."; // Rest of your query here