从表单中传递未编码的URL

时间:2017-05-26 08:34:44

标签: php forms url decode

我已在主页上设置了一个搜索框,该搜索框应通过传递网址链接到Opencart搜索。

我遇到的问题是生成的网址始终是未编码的:

<?php $url1 = json_decode(file_get_contents('http://www.movie.com:/retrieveMovies')); $counter = count($url1); for ($i = 0; $i < $counter; $i++){ $movieID = $url1[$i]->movieId ; $url2 = json_decode(file_get_contents(' http://movie.com/movieOne?seatNum=' . $movieID),TRUE); foreach ($url2 as $obj){ $testingResult = $obj['avail']; echo "<button onclick = testBTN('".$testingResult."')>Seat ID</button>"; } } ?> <script> function testBTN(x) { if (x == "OK"){ alert("Seat booked" ); } else { alert("Error in booking seats" ); } } </script>

我的目标是

store/index.php?route%3Dproduct%2Fsearch%26filter_name=test

我目前的代码是:

store/index.php?route=product/search&filter_name=test

我尝试过编码,解码,html_entity_decode,urldecode,rawurldecode ... 我已尝试将表单方法设为<?php $storesearch = $_REQUEST['store-search'] ; $filter = 'route=product/search&filter_name'; $filtered = html_entity_decode($filter); ?> <form id="store" method="GET" action="http://localhost/vinmobile/store/index.php?<?php echo $storesearch; ?>"> <input type="text" id="store-search" name="<?php echo $filtered; ?>" value="Store Search" onclick="this.value = '';"> <input type="submit" name="" value="Search"> </form> <?php echo $filtered; ?> // Just to show it onscreen to make sure it's written correctly 我已将变量声明尝试为GET, POST

每次,我仍然会看到这样的网址:

$_REQUEST, $_GET, $_POST

我确信这件事情我已经错过了,但我认为&#39;我已经尝试了一切,如果有人能指出我正确的方向,我将非常感激。

干杯

webecho

1 个答案:

答案 0 :(得分:0)

使用Javascript结束,它完美运行:

{{1}}

感谢你的建议