mysql_select_db("zaiping", $con);
$result = array();
$rs = mysql_query("select count(*) as count from dept where deptName like $deptName");
$row = mysql_fetch_array($rs);
ChromePhp::log($row);
$result["total"] = $row[0];
sql命令行执行select count(*) as count from dept where deptName like $deptName
是没有问题的,如果记录数>=2也是没问题的,问题就是当有1条匹配数据时$result["total"]=0,这是为什么,为什么不是1?谢谢
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