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我有一個php/Ajax/Jquery腳本，它將表單字段插入到MySQL中，並在您點擊提交時無需刷新即可更新頁面。我希望腳本提交四個表單字段，而不是一個。如何修改PHP/jquery/Ajax腳本有多個表單字段posst
我已經更新了數據庫表add_delete_record與另外3個領域：平衡，ACCOUNT_NUMBER和月，加上內容字段已經在那裏了。
下面可能是矯枉過正的代碼，因爲我只需要修改幾行，但我想這會回答所有問題。
這是PHP & html頁面：
<div class="content_wrapper">
<ul id="responds">
<?php
//include db configuration file
include_once("config.php");
//MySQL query
$Result = mysql_query("SELECT id,content FROM add_delete_record");
//get all records from add_delete_record table
while($row = mysql_fetch_array($Result))
{
echo '<li id="item_'.$row["id"].'">';
echo '<div class="del_wrapper"><a href="#" class="del_button" id="del-'.$row["id"].'">';
echo '<img src="images/icon_del.gif" border="0" />';
echo '</a></div>';
echo $row["content"].'</li>';
}
//close db connection
mysql_close($connecDB);
?>
</ul>
<div class="form_style">
<textarea name="content_txt" id="contentText" cols="45" rows="5"></textarea>
<button id="FormSubmit">Add record</button>
</div>
</div>
這是它張貼到PHP：
<?php
//include db configuration file
include_once("config.php");
//check $_POST["content_txt"] is not empty
if(isset($_POST["content_txt"]) && strlen($_POST["content_txt"])>0)
{
//sanitize post value, PHP filter FILTER_SANITIZE_STRING, FILTER_FLAG_STRIP_HIGH
$contentToSave = filter_var($_POST["content_txt"],FILTER_SANITIZE_STRING, FILTER_FLAG_STRIP_HIGH);
// Insert sanitize string in record
if(mysql_query("INSERT INTO add_delete_record(content) VALUES('".$contentToSave."')"))
{
//Record is successfully inserted, respond to ajax request
$my_id = mysql_insert_id(); //Get ID of last inserted record from MySQL
echo '<li id="item_'.$my_id.'">';
echo '<div class="del_wrapper"><a href="#" class="del_button" id="del-'.$my_id.'">';
echo '<img src="images/icon_del.gif" border="0" />';
echo '</a></div>';
echo $contentToSave.'</li>';
mysql_close($connecDB);
}else{
//output error
//header('HTTP/1.1 500 '.mysql_error());
header('HTTP/1.1 500 Looks like mysql error, could not insert record!');
exit();
}
}
elseif(isset($_POST["recordToDelete"]) && strlen($_POST["recordToDelete"])>0 && is_numeric($_POST["recordToDelete"]))
{//do we have a delete request? $_POST["recordToDelete"]
//sanitize post value, PHP filter FILTER_SANITIZE_NUMBER_INT removes all characters except digits, plus and minus sign.
$idToDelete = filter_var($_POST["recordToDelete"],FILTER_SANITIZE_NUMBER_INT);
//try deleting record using the record ID we received from POST
if(!mysql_query("DELETE FROM add_delete_record WHERE id=".$idToDelete))
{
//If mysql delete record was unsuccessful, output error
header('HTTP/1.1 500 Could not delete record!');
exit();
}
mysql_close($connecDB);
}else{
//Output error
header('HTTP/1.1 500 Error occurred, Could not process request!');
exit();
}
?>
這是JQuery的
$(document).ready(function() {
//##### Add record when Add Record Button is clicked #########
$("#FormSubmit").click(function (e) {
e.preventDefault();
if($("#contentText").val()==="") //simple validation
{
alert("Please enter some text!");
return false;
}
var myData = "content_txt="+ $("#contentText").val(); //post variables
jQuery.ajax({
type: "POST", // HTTP method POST or GET
url: "response.php", //Where to make Ajax calls
dataType:"text", // Data type, HTML, json etc.
data:myData, //post variables
success:function(response){
$("#responds").append(response);
$("#contentText").val(''); //empty text field after successful submission
},
error:function (xhr, ajaxOptions, thrownError){
alert(thrownError); //throw any errors
}
});
});
//##### Delete record when delete Button is clicked #########
$("body").on("click", "#responds .del_button", function(e) {
e.preventDefault();
var clickedID = this.id.split("-"); //Split string (Split works as PHP explode)
var DbNumberID = clickedID[1]; //and get number from array
var myData = 'recordToDelete='+ DbNumberID; //build a post data structure
jQuery.ajax({
type: "POST", // HTTP method POST or GET
url: "response.php", //Where to make Ajax calls
dataType:"text", // Data type, HTML, json etc.
data:myData, //post variables
success:function(response){
//on success, hide element user wants to delete.
$('#item_'+DbNumberID).fadeOut("slow");
},
error:function (xhr, ajaxOptions, thrownError){
//On error, we alert user
alert(thrownError);
}
});
});
});
這不是我的腳本所以我認爲我也應該給一個鏈接來信任它的作者： http://www.sanwebe.com/2012/04/ajax-add-delete-sql-records-jquery-php
感謝您的回答！我會照顧我所承諾的。我也會刪除以前的評論。 –
我注意到你將變量'$ contenttosave'的名稱改爲'$ content'，這很好，我只是好奇你設置'$ balance $ account $ monthly'的其他變量是否也是這樣假設在echo語句中？如果我假設添加這行代碼的3行if（isset（$ _ POST [「content_txt」]）&& strlen（$ _ POST [「content_txt」]）> 0）'3表單字段，你沒有名字？ –
這將取決於您對這兩個問題的要求。第一個問題：是的，如果你想要顯示的刪除網格包含新的值。第二個問題：是的，如果你想要3個新領域需要一個價值。如果你沒有問題，他們是空的，你不需要做驗證。 – erikrunia