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我想從同一頁面上的輸入填充從數據庫中的值下拉的相同頁面本身。使用PHP動態在同一頁面上填充下拉列表使用PHP
我在論壇上試過幾個選項,但沒有幫助。
點擊提交後會進入空白的下一頁。
我試過onchange = AjaxFunction();正如一篇文章中的建議,但我仍然得到一個空白頁。
任何幫助表示讚賞。
這是我form.php的
<form action="connection.php" class="form-solid-blue" method="get">
<div class="title">
<h2></h2>
<h2>Tracking & Receiving</h2></div>
<div class="element-input<?php frmd_add_class("input2"); ?>">
<label class="title"></label>
<div class="item-cont">
<input class="small" type="text" name="store" placeholder="Store #"/>
<span class="icon-place"></span>
</div>
</div>
<div class="element-input<?php frmd_add_class("input"); ?>">
<label class="title"></label>
<div class="item-cont">
<input class="medium" type="text" name="userid" placeholder="UserId"/>
<span class="icon-place"></span>
</div>
</div>
<div class="element-input<?php frmd_add_class("input1"); ?>">
<label class="title"></label>
<div class="item-cont">
<input class="large" type="text" name="order" placeholder="Order Number"/>
<span class="icon-place"></span>
</div>
</div>
<div class="submit">
<input type="submit" value="Send"/>
</div>
<div class="element-separator">
<hr>
<h3 class="section-break-title">Tracking Numbers</h3>
</div>
<div class="element-multiple<?php frmd_add_class("multiple"); ?>">
<label class="title"></label>
<div class="item-cont">
<div class="large">
<select data-no-selected="Nothing selected" name="multiple[]" multiple="multiple">
<option value="option_1">option 1</option>
<option value="option_2">option 2</option>
<option value="option_3">option 3</option>
</select>
<span class="icon-place"></span>
</div>
</div>
</div>
<div class="submit">
<input type="submit" value="Submit"/>
</div>
</form>
這是connection.php - 服務器端代碼
<?php
// Create connection to Oracle
$conn = oci_connect("XXXX", "xyxyx", "xyxyx");
if (!$conn) {
$m = oci_error();
echo $m['message'], "\n";
exit;
}
$query = "SELECT TRACKING_NUMBER FROM JC_SHIPPED_ORDER_TRACKING WHERE EXT_PURCHASE_ORDERS_ID = :order_bv";
$stid = oci_parse($conn, $query);
$order = $_GET['order'];
oci_bind_by_name($stid, ':order_bv', $order);
oci_execute($stid);
//Because order is a unique value I only expect one row
$row = oci_fetch_array($stid, OCI_ASSOC);
if (!$row) {
exit("The order " . $order . " is invalid. Please check and try again");
}
$trackID = $row['TRACKING_NUMBER'];
echo "<form name=form1 method=POST action='form.php'>";
//echo "<select name='TRACKING_NUMBER' onchange=AjaxFunction();>";
while ($row = oci_fetch_array($stid)) {
echo "<option value=\"option_1\">" . $row['TRACKING_NUMBER'] . "</option>";
}
echo "</select>";
//echo ("The order " . $order . " is valid.");
oci_free_statement($stid);
oci_close($conn);
?>
爲什麼你試試,如果你想以相同的形式攜手共創中端PHP文件另一種形式。 – PHJCJO
@PHJCJO - 感謝您的回覆。我試圖保持數據庫連接和查詢separate.And我想填充下拉form.php輸入和下拉的結果都在form.php – Max
是的,我明白了,但我看到這行回聲「