Question

我想从数据库中获取数据并将字符串添加到主题，然后将其放入另一个查询中。当我运行代码时，出现此错误：

致命错误：未捕获的PDOException：SQLSTATE [42000]：语法错误或访问冲突：1064您的SQL语法错误；检查与您的MySQL服务器版本相对应的手册在“ goldhyipPID”，“ goldhyipPayStatus”， programName，'goldhyipLastPayout'）值（在第1行的'1' C：\ wamp64 \ www \ allmonitors \ test.php在第69行

<?php
set_time_limit(3600);
require_once 'fetchdetails/func.php';
require_once 'config.php';
$stmt = $conn->prepare("SELECT * FROM `monitors`");
$stmt->execute();
$result = $stmt->fetchAll(PDO::FETCH_ASSOC);
foreach ($result as $monitor) {
    $monitorName = $monitor['monitorName'];
    $monitorNamePID = $monitorName . 'PID';
    $monitorNameLastPayout = $monitorName . 'LastPayout';
    $monitorNamePayStatus = $monitorName . 'PayStatus';
    $siteURL = $monitor['monitorurl'];
    $pattern1GetPID = $monitor['monitorPatternGetPID'];
    $patternLastPayOut = $monitor['monitorPatternLastPayout'];
    $patternPStatus = $monitor['monitorPatternPayStatus'];
    $patterndetailsurl = $monitor['monitorDetailsLink'];
    $patterngotositesurl = $monitor['monitorPatternGoSite'];
    $content = getPageContent($siteURL);
    preg_match_all($pattern1GetPID, $content, $matches, PREG_SET_ORDER, 0);
    foreach ($matches as $pid) {
        $id = $pid[1];
        $detailsurl = $patterndetailsurl . $id;
        $gositesurl = $patterngotositesurl . $id;
        $details = getPageContent($detailsurl);
        preg_match_all($patternPStatus, $details, $status);
        $payingStatusNumber = $status[1][0];
        if ($payingStatusNumber == 4) {
            $payingStatus = 'Not Paying';
        } elseif ($payingStatusNumber == 3) {
            $payingStatus = 'Problem';
        } elseif ($payingStatusNumber == 2) {
            $payingStatus = 'Waiting';
        } elseif ($payingStatusNumber == 1) {
            $payingStatus = 'Paying';
        }
        preg_match_all($patternLastPayOut, $details, $payout);
        if (isset($payout[1][0])) {
            $payoutdate = $payout[1][0];
        } else {
            $payoutdate = ' Not Set';
        };
        $stmt2 = $conn->prepare('SELECT * FROM programs where :monitorNamePID=:id');
        $stmt2->bindParam('monitorNamePID', $monitorNamePID);
        $stmt2->bindParam('id', $id);
        $stmt2->execute();
        $numofupdates = $stmt2->rowCount();
        if ($numofupdates >= 1) {
            $stmt3 = $conn->prepare("UPDATE programs SET :monitorNamePayStatus=:payingstatus , :monitorNameLastPayout=:goldhyiplastpayout WHERE :monitorNamePID=:goldhyippid  ");
            $stmt3->bindParam('monitorNamePayStatus', $monitorNamePayStatus);
            $stmt3->bindParam('monitorNameLastPayout', $monitorNameLastPayout);
            $stmt3->bindParam('monitorNamePID', $monitorNamePID);
            $stmt3->bindParam('payingstatus', $payingStatus);
            $stmt3->bindParam('goldhyiplastpayout', $payoutdate);
            $stmt3->bindParam('goldhyippid', $id);
            $stmt3->execute();
            echo 'P ID Updated : ' . $id . '<br>';
        } else {
            $siteAddress = get_redirect_final_host_url($gositesurl);
            echo $siteAddress;
            $stmt3 = $conn->prepare('INSERT INTO programs (:monitorNamePID, :monitorNamePayStatus, programName,:monitorNameLastPayout) VALUES (:goldhyippid, :payingstatus, :progname, :goldhyiplastpayout)');
            $stmt3->bindParam('monitorNamePID', $monitorNamePID);
            $stmt3->bindParam('monitorNamePayStatus', $monitorNamePayStatus);
            $stmt3->bindParam('monitorNameLastPayout', $monitorNameLastPayout);
            $stmt3->bindParam('goldhyippid', $id);
            $stmt3->bindParam('payingstatus', $payingStatus);
            $stmt3->bindParam('progname', $siteAddress);
            $stmt3->bindParam('goldhyiplastpayout', $payoutdate);
            $stmt3->execute();
            echo 'P ID inserted : ' . $id . '<br>';
        }
    }
    echo "Fetching $siteURL Done <br>";
}

我从sql中获取了详细信息，并向其中添加了字符串，用于PDO查询中。当我写列名没问题，但是当我使用变量时，我得到了错误。

Answer 1

列名不应以冒号开头。在:和UPDATE查询中，从列名中删除冒号INSERT。绑定表（或列）名称无效。

$stmt3 = $conn->prepare("UPDATE programs SET monitorNamePayStatus=:payingstatus , monitorNameLastPayout=:goldhyiplastpayout WHERE monitorNamePID=:goldhyippid  ");

$stmt3 = $conn->prepare('INSERT INTO programs (monitorNamePID, monitorNamePayStatus, programName, monitorNameLastPayout) VALUES (:goldhyippid, :payingstatus, :progname, :goldhyiplastpayout)');

Answer 2

绑定表（或列）名称无效

SQLSTATE [42000]：语法错误或访问冲突：1064在bindparam中设置变量时出错

2 个答案: