sql UPDATE语句因SQL错误而失败

Question

继续收到此警告:

警告:odbc_exec():SQL错误:[Microsoft] [ODBC Microsoft Access驱动程序]查询表达式''WHERE id = 86'中的字符串语法错误.,SQLExecDirect中的SQL状态37000

无论我如何编辑此SQL UPDATE命令:

$sqlU = "UPDATE inv20152016
        SET coNo = '$coNo', cusNo = '$cusNo', rcvTech = '$rcvTech', 
            rcvLoc = '$rcvLoc', rcvDate = '$rcvDate', rcyTech = '$rcyTech', 
            rcyLoc = '$rcyLoc', rcyDate = '$rcyDate', rtnTech = '$rtnTech', 
            rtnLoc = '$rtnLoc', rtnDate = '$rtnDate', cusInvNo = '$cusInvNo', 
            notes = '$notes
        WHERE id = '$rcyId'";
    if (!$rs=odbc_exec($conn,$sqlU)) {
        echo '<br />Error UPDATE\'ing new record: (' . $sqlU . ')<br />';
        goto end;
    } else {
        echo '<br />RECORD successfully UPDATE\'d in database.recycle20152016<br />'; //$sqlUpdate<br />;
        header('location: searchDb.php');
        exit();
    }

尝试提交UPDATE后,我在提交页面上返回以下内容:

错误更新新记录:(更新inv20152016 SET coNo ='658454',cusNo ='3282',rcvTech ='Dave Phillips',rcvLoc ='Sparington',rcvDate ='2016-11-15',rcyTech ='Melo Hazak',rcyLoc ='Wildmere',rcyDate ='2016-02-16',rtnTech ='Dave Phillips',rtnLoc ='Sparington',rtnDate ='2016-02-25',cusInvNo ='',notes =' WHERE id = 86)

似乎所有UPDATE变量都符合预期,包括idWHERE子句中的变量.我玩过vars之间的引用和间距没有变化.我已经验证了ACCESS后端中的数据类型与推送更新的数据相匹配.

任何想法为什么这个UPDATE语句可能无法正常工作？

Answer 1

你'从notes = '$notescolumnId var中留下一个引号.

尝试:

$sqlU = "UPDATE inv20152016
        SET coNo = '$coNo', cusNo = '$cusNo', rcvTech = '$rcvTech', rcvLoc = '$rcvLoc', rcvDate = '$rcvDate', rcyTech = '$rcyTech', rcyLoc = '$rcyLoc', rcyDate = '$rcyDate', rtnTech = '$rtnTech', rtnLoc = '$rtnLoc', rtnDate = '$rtnDate', cusInvNo = '$cusInvNo', notes = '$notes'
        WHERE id = '$rcyId'";
    if (!$rs=odbc_exec($conn,$sqlU)) {
        echo '<br />Error UPDATE\'ing new record: (' . $sqlU . ')<br />';
        goto end;
    } else {
        echo '<br />RECORD successfully UPDATE\'d in database.recycle20152016<br />'; //$sqlUpdate<br />;
        header('location: searchDb.php');
        exit();
    }