sql - Masalah nombor rekod pertanyaan PHP, pengecualian kiraan

Question

    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 command line executionselect count(*) as count from dept where deptName like $deptName tiada masalah Jika bilangan rekod >= 2, masalahnya ialah apabila terdapat 1 data yang sepadan, $result["total"]=0 Kenapa bukan 1? Terima kasih

Answer 1

你把$row[0]改成$row[0]['count']试试;显示过滤了

Answer 2

$rs = mysql_query("select count(*) as count from dept where deptName like '%".$deptName."%'");
SQL like 请加百分号

Answer 3

Bila masalah macam ni berlaku, selalunya saya selesaikan macam ni:

$sql = "select count(*) as count from dept where deptName like $deptName";
print_r($sql);        //将输出的SQL拿到MySQL去执行，看报什么错误
$re = mysql_query($sql);
var_dump($re);        //查看返回什么

Berdasarkan output, buat pertimbangan yang sepadan untuk melihat di mana masalahnya.