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[英]Submit a form and display text box value in another tab in AJAX & PHP
[英]How to fill a form input text box based on the value of another text box using Ajax
我正在學習php,html和ajax。 我已經建立了一個包含員工信息的MySQL數據庫。 我設法弄清楚了如何使用Ajax自動填充文本框(稱為“員工詳細信息”)。 當您開始輸入員工的姓名時,它將由他們的全名和公司名稱組成。
我現在想要做的是根據第一個文本框的值自動用其員工ID填充第二個文本框。
我搜索了很多問題和教程,但是找不到關於如何執行此操作的簡單說明,我發現的示例完全沒有包括php,ajax,html,而且我不知道該怎么做全部結合在一起(我不是編碼天才,我無法使用任何示例)。 我已經在這個問題上停留了幾個小時,失去了生存的意願!
如果有人可以在一個地方提供一個簡單的解釋,例如php,ajax和html的示例,那么我將不勝感激!
到目前為止,這是我的代碼。
form.php
<link rel="stylesheet" href="//code.jquery.com/ui/1.11.4/themes/smoothness/jquery-ui.css">
<script src="//code.jquery.com/jquery-1.10.2.js"></script>
<script src="//code.jquery.com/ui/1.11.4/jquery-ui.js"></script>
<script>
$(function() {
$( "#employeedetails" ).autocomplete({
source: 'search.php'
});
});
</script>
<div class="form-group">
<b class="text-primary">Employee details:</b>
<input type="text" class="form-control" value="" id="employeedetails" name="employeedetails" required>
<b class="text-primary">Id:</b>
<input type="text" name="employeeid" id="employeeid" placeholder="employeeid"/>
</div>
search.php
include 'dbconnect.php';
//get search term
$searchTerm = $_GET['term'];
//get matched data from employee table
$query = $conn->query("SELECT *
FROM employees
WHERE (firstname LIKE '%".$searchTerm."%')
OR (surname LIKE '%".$searchTerm."%')
OR (companyname LIKE '%".$searchTerm."%')
ORDER BY firstname ASC
");
while ($row = $query->fetch_assoc()) {
$data[] = $row['firstname'] . " " . $row['surname'] . " - " .
}
//return json data
echo json_encode($data);
?>
假設您的HTML員工詳細信息文本框具有id =“ employeeDetail”,而您的HTML員工ID文本框具有id =“ employeeId”,則您可以使用jQuery來監聽對員工詳細信息的選擇,然后通過Ajax發送該選擇,並使用Ajax響應更新您的員工ID文本框的值。 這將涉及以下jQuery和PHP代碼:
jQuery代碼:
$(document).ready(function(){
$(#employeeDetail).on("change", function(){ //use an appropriate event handler here
$.ajax({
method: "POST",
url: "getEmployeeId.php",
data: {
employee_detail: $("#employeeDetail").val(),
},
success: function(response){
if (response == "FALSE") {
var message = "ERROR: something went wrong on the MYSQL side";
alert(message);
} else {
$("#employeeId").val(response);
}
},
error: function(jqXHR, textStatus, errorThrown){
var message: "ERROR: something went wrong with the AJAX call - " + textStatus + " - " + errorThrown;
alert(message);
}
});
});
});
PHP代碼(getEmployeeId.php):
//set $server, $user, $password and $database_name, then establish the connection:
$conn = new mysqli($server, $user, $password, $database_name);
if ($conn->connect_error) {
exit("FALSE");
}
//get employee detail from POST (sent via AJAX):
$employee_detail = $_POST["employee_detail"];
//here, you should test whether employee_detail matches what you expect
//here, split $employee_detail into $first_name, $last_name and $company_name
//now you are ready to send the MYSQL query:
$sql = 'SELECT employeeid FROM tablename WHERE firstname = "$first_name" AND surname = "$last_name" AND companyname = "$company_name"';
//since you expect a single matching result, you can test for num_rows == 1:
if ((! $result = $_conn->query($sql)) || ($result->num_rows !== 1)) {
exit("FALSE");
} else {
while ($row = $result->fetch_assoc()) {
$employee_id = $row['id'];
}
}
//now send $employee_id back to the AJAX call:
echo $employee_id;
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