表单在php中提交两次

Question

我分别为新患者和已经登记的患者提供两种约会表格，手机号码作为主要密码。

因此，在新患者登记中，我们检查该手机是否已经存在，如果是，则显示错误，否则在表中输入新数据。

但是目前，即使我输入一个新的手机号码，它也会显示“已经存在”错误，并且还将数据输入db。

所以我认为该表单已提交了2次，但是我不知道发生了什么。

我的php文件代码段是：

if ($_POST['isnewpatient'] == "true") {

    @$name = mysql_real_escape_string(trim($_POST['aaptntname']));
    @$email = mysql_real_escape_string(trim($_POST['emlid']));
    @$mobile = mysql_real_escape_string(trim($_POST['mobile']));
    $qqcSql = "select * from " . WP_eemail_TABLE_SUB 
      . " where eemail_mobile_sub ='" . trim($_POST['mobile']) 
      . "' OR eemail_patient_id ='" . trim($_POST['mobile']) . "'";

    $qqdata1 = $wpdb->get_results($qqcSql);
    var_dump($qqdata1);

    if (!empty($qqdata1)) {

        $err = 1;
        echo "<div id='message' class='aerror'>Already patient details exists. Use your existing patient ID !</div>\n";

    } 
    else {
        $pt_id = mysql_real_escape_string(trim($_POST['mobile']));
        $sql = "insert query to WP_Appointment";

        $wpdb->get_results($sql);

        $sqls = "insert query to WP_Appointment_Contact";

        $wpdb->get_results($sqls);

        $sqql = " insert query to table WP_eemail_Table_Sub";

        $wpdb->get_results($sqql);

        echo "<div id='message' class='asuccess'>Request has been sent for appointment  </div>";

    }
} 
 else {

   // Already registered patient form

 }

<form name="FormEdit" action="<?php echo the_permalink(); ?>" method="post" onsubmit="return p_appointment()" class="aform">

   /* Form part */  
</form>

JavaScript部分

function p_appointment()
{ 
    if($('input:radio[name=new_patient]:checked').val() == "new") 
        document.FormEdit.isnewpatient.value = "true";
    if($('input:radio[name=new_patient]:checked').val() == "old") 
        document.FormEdit.isnewpatient.value = "false";  

    document.FormEdit.appsmssend.value = "true";
    document.FormEdit.appemailsend.value = "true";

}

Answer 1

将您的js方法更改为此：

function p_appointment()
{ 
    if($('input:radio[name=new_patient]:checked').val() == "new") 
        document.FormEdit.isnewpatient.value = true;
    if($('input:radio[name=new_patient]:checked').val() == "old") 
        document.FormEdit.isnewpatient.value = false;  


  // document.FormEdit.appsmssend.value = "true";
   // document.FormEdit.appemailsend.value = "true"; commented because i did not added these field while testing.
    return document.FormEdit.isnewpatient.value;

}

这是我用作形式：

<form name="FormEdit"  method="post" onsubmit="return p_appointment()" class="aform">
<input type="text" name="name"/>
<input type="submit" value="submit"/>
</form>

不使用：

if ($_POST['isnewpatient'] == "true") {

}
if ($_POST['isnewpatient'] == "false") {

}

采用：

if($_POST['isnewpatient']) {

}else {

}