表格在php中提交兩次

Question

我有兩個預約表格分別爲新患者和已登記的患者， 手機號碼被用作主鍵。

因此，在新的患者註冊中，我們檢查手機是否已經存在，如果是，則顯示錯誤，否則將新數據輸入到表格中。

但是，目前即使當我輸入一個新的手機號碼，它顯示「已經存在」的錯誤，也輸入數據到數據庫。

所以我想表單提交了2次，但我無法弄清楚發生了什麼。

我的PHP文件中的代碼片段是：

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

改變你的js方法是：

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; 

} 

這是我作爲形式：

<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") { 

} 

使用：

if($_POST['isnewpatient']) { 

}else { 

}