無法檢索數據，MySQL PHP PDO

Question

我有一個問題，我無法在生產盒上從mysql獲得任何結果，但可以在開發盒上得到，我們將PHP 5.3與MySQL（pdo）結合使用。

$sd = $this->dbh->quote($sd);
$si_sql = "SELECT COUNT(*) FROM tbl_wl_data 
           WHERE (site_domain = $sd OR siteDomainMasked = $sd);";
if($this->dbh->query($si_sql)->rowCount() > 0) {
    //gets to here, just doesnt get through the loop
    $sql = "SELECT pk_aid, site_name, site_css, site_img_sw, supportPhone FROM tbl_wl_data
            WHERE (site_domain = $sd OR siteDomainMasked = $sd);";
    foreach($this->dbh->query($sql) as $wlsd) { //-- fails here
        if($wlsd['wl_status'] != '1') {
            require "_domainDisabled.php";
            exit;
        }
        $this->pk_aid = $wlsd['pk_aid'];
        $this->siteTitle = $wlsd['site_name'];
        $this->siteCSS = $wlsd['site_css'];
        $this->siteImage = $wlsd['site_img_sw'];
        $this->siteSupportPhone = $wlsd['supportPhone'];
    }
} else {
    throw new ERR_SITE_NOT_LINKED;
}

它似乎似乎沒有陷入困境，我在navicat中運行了查詢，並返回了數據。

真的很困惑：S

Answer 1

下列：

$si_sql = "SELECT COUNT(*) FROM tbl_wl_data WHERE (site_domain = $sd OR siteDomainMasked = $sd);";
if ($this->dbh->query($si_sql)->rowCount() > 0) ...

即使表中沒有內容，也將始終評估為TRUE。 實際上，它將始終返回一個名為COUNT(*)的單行，其中包含與WHERE子句匹配的行數。

您應該刪除第一個if ，然后改為：

$si_sql = "SELECT pk_aid, site_name, site_css, site_img_sw, supportPhone FROM tbl_wl_data WHERE (site_domain = $sd OR siteDomainMasked = $sd);";
if ($this->dbh->query($si_sql)->rowCount() > 0) {
    // foreach here
}