我有一個頁面,有搜索欄。該頁面被設計爲在提交鏈接被點擊時在具有ID的同一頁面上得到結果,結果將出現在下面。提交按鈕不是傳統的提交按鈕。我的問題是,我已經做了所有的PHP的東西,但是當我將讓結果出現在點擊鏈接時,PHP不運行。有人能幫助我嗎?下面是代碼。謝謝通過點擊鏈接的href提交表單,而無需使用一個提交按鈕
<form action="" method="post">
<div class="col-md-9">
<div class="row">
<div class="col-md-7 no-padding">
<div class="inner">
<h3>Track & Trace</h3>
<span>Already have a load ID, please insert it below</span>
</div>
</div>
<div class="col-md-5 no-padding">
<input class="input-fullwidth" name="track" id="track">
</div>
</div>
</div>
<div class="col-md-3">
<a id="track-it" href="#section-tracking-result" class="btn-custom btn-fullwidth">Track it</a>
</div>
</form>
<?php
define('DB_HOST','localhost');
define('DB_USER','username');
define('DB_PASS','password');
define('DB_NAME','dbname');
$conn = mysql_connect(DB_HOST,DB_USER,DB_PASS) or die(mysql_error());
mysql_select_db(dbname) or die(mysql_error());
$output = '';
if(isset($_POST['track'])) {
if(empty($_POST['track'])) {
echo "Tracking Code is Required";
}
$searchq = $_POST['track'];
$searchq = preg_replace("#[^0-9a-z]#i", "",$searchq);
$query = mysql_query("SELECT * FROM track WHERE keywords LIKE '%$searchq%'");
$count = mysql_num_rows($query);
if($count == 0) {
$output = 'There was no search results!';
} else {
while($row = mysql_fetch_array($query)) {
$consignee = $row['consignee'];
$desti = $row['desti'];
$date = $row['date'];
$id = $row['id'];
}
}
}
?>
它似乎並不像你甚至有
照常使用表格;創建提交表單的鏈接並使用onclick =「form.submit();」鏈路 – Sinto