php - 如何更新 3 个表列，而页面只有一列可见

Question

这是一个表列，显示消息取决于用户在数据库中的首选语言选择。例如，如果首选语言是 ENG 只有 eng消息将显示PS:在数据库中，有3列存储了3种不同语言的消息

 <?php
 $possibleLang = ["繁體","简体","ENG"];
 $testAreaField = ["traditionalmessage","simplifiedmessage","engmessage"];
 $treatmentName = ["vaccinename1","vaccinename2","vaccinename3"];
 $treatmentNameSuffix = ["\n下一個注射期為:","\n下一个注射期为:","\nNext 
 injection period will be:"];

 $index = array_search($row['language'],$possibleLang);
 ?>
<td>
<textarea  rows="3" cols="18" class="url" name="<?php echo 
 $testAreaField[$index]; ?>[]" data-value="<?php echo 
 $row[$treatmentName[$index]] . $treatmentNameSuffix[$index]; ?>"><?php echo 
 $row[$testAreaField[$index]]; ?>
 </textarea>
 </td>

此表位于显示来自数据库的数据并将数据更新回数据库的编辑页面。

 if (isset($_POST['submit'])){
  foreach ($_POST['patientid'] as $index => $patientid) {
    $id = mysql_real_escape_string($_POST['id'][$index]);
    $data1 = mysql_real_escape_string($patientid);
    $data17 = mysql_real_escape_string($_POST['patienthkid'][$index]);

    $data2 = mysql_real_escape_string($_POST['vaccineid'][$index]);
    $data5 = mysql_real_escape_string($_POST['vaccinename3'][$index]);
    $data6 = mysql_real_escape_string($_POST['totalnoofinjection'][$index]);
    $data7 = mysql_real_escape_string($_POST['nthinjection'][$index]);
    $data8 = mysql_real_escape_string($_POST['date'][$index]);
    $data9 = mysql_real_escape_string($_POST['nextdate'][$index]);
    $data10 = mysql_real_escape_string($_POST['skip'][$index]);
    $data11 = mysql_real_escape_string($_POST['language'][$index]);
    $data12 = mysql_real_escape_string($_POST['traditionalmessage'][$index]);
    $data13 = mysql_real_escape_string($_POST['simplifiedmessage'][$index]);
    $data14 = mysql_real_escape_string($_POST['engmessage'][$index]);
    $data15 = mysql_real_escape_string($_POST['status'][$index]);
    $data16 = mysql_real_escape_string($_POST['nurse'][$index]);


   if ($data12 != null) {
    mysql_query("UPDATE patientvaccinedetail SET patientid 
   ='$data1',patienthkid 
   ='$data17',vaccineid='$data2',vaccinename3='$data5', 
    totalnoofinjection='$data6',nthinjection='$data7',
    date='$data8',nextdate='$data9',skip='$data10',language='$data11',
    traditionalmessage='$data12',status='$data15',nurse='$data16' WHERE 
    id=$id")   or die(mysql_error());
     }
    else if ($data13 != null) {
   mysql_query("UPDATE patientvaccinedetail SET patientid 
    ='$data1',patienthkid 
   ='$data17',vaccineid='$data2',vaccinename3='$data5', 
  totalnoofinjection='$data6',
    nthinjection='$data7',date='$data8',nextdate='$data9',
   skip='$data10',language='$data11',
   simplifiedmessage='$data13',status='$data15',nurse='$data16' WHERE 
    id=$id")   or die(mysql_error());
    }
  else if ($data14 != null) {
  mysql_query("UPDATE patientvaccinedetail SET patientid 
  ='$data1',patienthkid ='$data17',vaccineid='$data2',
  vaccinename3='$data5', totalnoofinjection='$data6',
    nthinjection='$data7',date='$data8',nextdate='$data9',
   skip='$data10',language='$data11',
   engmessage='$data14',status='$data15',nurse='$data16' WHERE id=$id")
   or die(mysql_error());
    }

    }
    }

问题是每当我单击“保存”时，消息的只有一个列被保存，其他两列在数据库中更新为 NULL。期望的结果是同时更新 3 列，而页面只有 1 列可见。任何想法。提前致谢

Answer 1

不要更新不是更新目标的列。例如，如果选择了传统消息，那么您将只更新此列。我可能会在这里使用单独的查询。

foreach ($_POST['patientid'] as $index => $patientid) {
    $id = mysql_real_escape_string($_POST['id'][$index]);
    $data1 = mysql_real_escape_string($patientid);
    $data2 = mysql_real_escape_string($_POST['traditionalmessage'][$index]);
    $data3 = mysql_real_escape_string($_POST['simplifiedmessage'][$index]);
    $data4 = mysql_real_escape_string($_POST['engmessage'][$index]);

    if ($data2 != null && $data2 != '') {
        mysql_query("UPDATE table SET patientid ='$data1',
          traditionalmessage='$data2' WHERE id=$id")   or die(mysql_error());
    }
    else if ($data3 != null && $data3 != '') {
        mysql_query("UPDATE table SET patientid ='$data1',
          simplifiedmessage='$data3' WHERE id=$id")   or die(mysql_error());
    }
    else if ($data4 != null && $data4 != '') {
        mysql_query("UPDATE table SET patientid ='$data1',
          engmessage='$data4' WHERE id=$id")   or die(mysql_error());
    }
}

请注意，我相信您正在为 MySQL 使用非常古老且已弃用的 PHP API。对于初学者来说，使用准备好的语句会让你受益匪浅。