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誰かが助けてくれるかどうか疑問に思っていました（この主題が以前に提起されたことを知って申し訳ありません）

サインアップ フォームを提出する前に、会社と電子メール アドレスが存在しないことを確認したい。

どんな助けにも感謝します。

テーブル名は R_TABLE で、フィールドは COMPANY です テーブル名は U_TABLE で、フィールドは EMAIL です

    <?
    session_start();
    //ignore this -->$email = $_POST['email'] ;

    //error_reporting(E_ALL); 
    //ini_set("display_errors", 1); 

    //Connect to Database
    include('includes/dbconn.php');
    ?>
    <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN"    "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
   <html xmlns="http://www.w3.org/1999/xhtml">
   <head>
   <meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" />
   <title>Sign Up Form</title>
   <link rel="shortcut icon" href="favicon.ico" />



   <link href="eps.css" rel="stylesheet" type="text/css" />
   <div class="header2">
   </div>

   //VALIDATES FIELDS

     <script language="javascript">

     function validateForm()
     {

 var x=document.forms["form1"]["email"].value;
 var atpos=x.indexOf("@");
 var dotpos=x.lastIndexOf(".");
 if (atpos<1 || dotpos<atpos+2 || dotpos+2>=x.length)
 {
    alert("Not a valid e-mail address");
    return false;
 } 
if (document.forms["form1"]["email"].value != document.forms["form1"]  ["email2"].value) {
    alert("Email addresses do not match");
    return false;
 }


    {
    var x=document.forms["form1"]["companyName"].value;
    if (x==null || x=="")
    {
    alert("Please enter your company name.");
    return false;
    }




    }
    }

    </script>






    </head>


    <body background="images/background.gif" repeat >
    <table width="1101" border="0" align="center" cellpadding="0" cellspacing="0">
    <tr>
    <td>
    <br/> <br/>



 <p align="center" class="SignUpText">Sign Up Form</p>
     <table width="600" border="0" align="center" cellpadding="0" cellspacing="0" class="TableBorder">
     <tr>
     <td>
     <form id="form1" name="form1" method="post" action="signup.php" onSubmit="return validateForm();">
    <input name="a" type="hidden" id="a" value="signup" />
    <input name="group" type="hidden" id="group" />

    <table width="100%" border="0" align="center" cellpadding="5" cellspacing="0" class="MainText">
    <tr>
    <td colspan="4" class="auto-style4" style="height: 40px"><strong>
<span class="auto-style5">Personal Information</span><br/>
    </tr>

     //COMPANY NAME

    <tr>
    <td width="25%">Company Name</td>
    <td style="width: 2%">*</td>
    <td width="25%"><label>

 <input name="companyName" type="text" class="auto-style3" id="companyName" size="35" /></label></td>

     <td style="width: 2%"><a href="#" class="hintanchor" onMouseover="showhint('Please enter your registered company name.', this, event, '150px')">?</a></td>


      //EMAIL

      <td>
  <input name="email" type="text" class="auto-style3" id="email" size="40" /></td>
      <tr>
      <td>Re-enter Email Address</td>
      <td style="width: 2%">*</td>
      <td>
  <input name="email2" type="text" class="auto-style3" id="email2" size="40" />    </td>

      //SUBMIT

      <tr>
      <td colspan="4"><div align="center">
      <input type="submit" name="Submit" value="Sign Up" />
      </div>
      </td>
      </tr>
      </table>
      </form>
      </td>
      </tr>
      </table>


Here is part of the class that adds new users, the email actually is added to contactEmail field , company is added to companyname within the database which I need to check if it doesn't already have the email and company name stored

function addmember($companyname,  $licencenumber,$companyNumber,$vatNumber,$address1,$address2,$address3,$town,$county,$postcode,$firstname,$surname,$email,$phone,$training,$norepairs,$logo,$repairtype,$repairgroup,$mode)
{

    $ref = "REP";
    $licencenumber = "TEST";

    $query = "INSERT INTO r_table VALUES ";
    $query.= "(NULL,";
    $query.= "\"$ref\",";
    $query.= "\"$companyname\",";
    $query.= "\"$licencenumber\",";
    $query.= "\"$companyNumber\",";
    $query.= "\"$vatNumber\",";
    $query.= "\"$address1\",";
    $query.= "\"$address2\",";
    $query.= "\"$address3\",";
    $query.= "\"$town\",";
    $query.= "\"$county\",";
    $query.= "\"$postcode\",";
    $query.= "\"$firstname\",";
    $query.= "\"$surname\",";
    $query.= "\"$email\",";
    $query.= "\"$phone\",";
    $query.= "\"$training\",";
    $query.= "\"$norepairs\",";
    $query.= "\"$logo\",";
    $query.= "\"$repairtype\",";
    $query.= "\"$repairgroup\",";
    $query.= "NOW(),";
    $query.= "NOW(),";
    $query.= "$mode)";

    $result = mysql_query($query);

    $insertid = mysql_insert_id();
    $ref = "REP".$insertid;
    $licencenumber = "test".$insertid;

    $query = "UPDATE r_table SET ref = \"$ref\", licenseNumber=\"$licencenumber\" WHERE id = ".$insertid;

    $result = mysql_query($query);

    echo mysql_error();

    if (!$result)
        echo mysql_error();
    else
        return $ref;
4

1 に答える 1

1

以下のようなことを 1 つ行うことができます。

$query = "select * from `tablename` where `fieldname` = 'value'";
$result = mysql_query($query) or die(mysql_error());
$num_result = mysql_num_rows($result);
if($num_result > 0){
    //Value exist for field already
 }

ここでmysql_num_rows戻りnumber of rows of result of queryます。これを使用すると、会社と電子メールが既に存在するかどうかを確認できます。

于 2013-01-16T13:26:45.500 に答える