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我正在使用Javascript創建一個PHP頁面。我創建了一個HTML表單，並將其稱爲代碼的PHP部分。當我顯示JSON數組時，我可以看到顯示的元素一毫秒，然後頁面刷新到初始頁面。代碼有效，但我不想自動刷新。我怎麼做？Html頁面自動刷新
HTML：
<div data-role="collapsible">
<h1>ADD INVNTORY</h1>
<form action='<?php echo htmlspecialchars($_SERVER["PHP_SELF"]);?>' method="post">
<fieldset>
<legend>
<span class="number">1</span>
Your basic info
</legend>
<label for="name">Name:</label>
<input type="text" id="name" name="product_name"/>
<label for="serial">Serial number:</label>
<input type="text" id="serial" name="serial"/>
<label for="amount">Amount:</label>
<input type="number" id="amount" name="amount"/>
<input type="submit" name="submit1" value="Submit"/>
</fieldset>
</form>
</div>
<form action="<?php echo htmlspecialchars($_SERVER["PHP_SELF"]);?>" method="post">
<div data-role="collapsible">
<h1>VIEW ALL</h1>
<fieldset>
<legend>
<span class="number">1</span>
Current Inventory list
</legend>
<input type="submit" name="submit2" value="Show all">
<div data-role="popup" id="myPopup" class="ui-content">
<h1 id="here"></h1>
</div>
</fieldset>
</div>
<div data-role="collapsible">
<h1>VIEW INVENTORY DETAILS(GRAPH)</h1>
<fieldset></fieldset>
</div>
<button type="submit">Sign Up</button>
</form>
PHP：
<?php
if (isset($_POST['submit1'])) {
$nname = "";
$sserial = "";
$aamount = "";
if ($_SERVER["REQUEST_METHOD"] == "POST") {
if (empty($_POST["product_name"])) {
{
echo "Name is required";
break;
}
} else {
$nname = test_input($_POST["product_name"]);
}
if (empty($_POST["serial"])) {
{
echo "Serial code is required";
break;
}
} else {
$sserial = test_input($_POST["serial"]);
}
if (empty($_POST["amount"])) {
{
echo "Amount is required";
break;
}
} else {
$aamount = test_input($_POST["amount"]);
}
insertdata($nname,$sserial,$aamount);
}
}
function test_input($data) {
return $data;
}
function insertdata($nname,$sserial,$aamount) {
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "mydb";
$conn = new mysqli($servername, $username, $password, $dbname);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "INSERT INTO inventory_list (name, serialno, amount) VALUES ('$nname', '$sserial', '$aamount')";
if ($conn->query($sql) === TRUE) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
$conn->close();
}
if (isset($_POST['submit2'])) {
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "mydb";
$conn = new mysqli($servername, $username, $password, $dbname);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT * from inventory_list";
$result = $conn->query($sql);
$return_arr=array();
if ($result->num_rows > 0) {
// output data of each row
while ($row=$result->fetch_assoc()) {
$row_array['name'] = $row['name'];
$row_array['serialno'] = $row['serialno'];
$row_array['amount'] = $row['amount'];
array_push($return_arr, $row_array);
}
} else {
echo "0 results";
}
$conn->close();
}
?>
^^ PHP的一部分，
的Javascript：
var arr = <?php echo json_encode($return_arr);?>, out = "<table>";
for (var i = 0; i < arr.length; i++) {
out += "<tr><td>" +
arr[i].name +
"</td><td>" +
arr[i].serialno +
"</td><td>" +
arr[i].amount +
"</td></tr>";
}
out += "</table>";
document.querySelector("#myPopup #here").innerHTML = out;
有上面沒有代碼可以刷新頁面 –
你的地方trigging表單提交您所提供的代碼之外？ –
正如@Dagon所說，這裏沒有任何東西可以引起你所描述的行爲。 –