AJAX无法发送到PHP
发布时间:2020-12-16 02:47:24 所属栏目:百科 来源:网络整理
导读:我试图将表单信息从index.html发送到output.php(两者都在同一目录中),但我收到此错误: Warning: Unknown: failed to open stream: No such file or directory in Unknown on line 0Fatal error: Unknown: Failed opening required 'C:/xampp/htdocs/Tese_Jo
我试图将表单信息从index.html发送到output.php(两者都在同一目录中),但我收到此错误:
Warning: Unknown: failed to open stream: No such file or directory in Unknown on line 0 Fatal error: Unknown: Failed opening required 'C:/xampp/htdocs/Tese_Jo?o/test-searchMYSQL/output.php' (include_path='.;C:xamppphpPEAR') in Unknown on line 0 如果您也可以帮助我,使用此代码我只能通过按ENTER或在搜索框外单击而不是只在搜索框中写入来获得AJAX的优势,任何人都知道如何解决这个问题? index.html的: <!DOCTYPE html> <html> <head> <title> Escolha de molecula</title> <script> function showUser(str) { if (str=="") { document.getElementById("ajax").innerHTML=""; return; } if (window.XMLHttpRequest) { // code for IE7+,Firefox,Chrome,Opera,Safari xmlhttp=new XMLHttpRequest(); } else { // code for IE6,IE5 xmlhttp=new ActiveXObject("Microsoft.XMLHTTP"); } xmlhttp.onreadystatechange=function() { if (xmlhttp.status==200) { document.getElementById("ajax").innerHTML=xmlhttp.responseText; } } var txq=document.getElementById("textquery").value; xmlhttp.open("POST","output.php?",true); xmlhttp.setRequestHeader("Content-type","application/x-www-form-urlencoded"); xmlhttp.send("textquery=" + txq); } </script> </head> <body> <h1 style="text-align:center;font-size: 60px"> Search for molecule in our database </h1> <hr> <p style="color:blue"> <q> I have lived much of my life among molecules. They are good company. </q> George Wald </p> <hr> <p style="font-family:verdana"> just write something in the search bar and it will retrieve it <br> More information at: <a href="http://xldb.fc.ul.pt/" target="_blank">http://xldb.fc.ul.pt/</a> </p> search: <input type="text" id="textquery" onchange="showUser()" ><br> </form> <br> <div id="ajax"><b>search results are displayed here</b></div> </body> </html> output.php: <?php $con=mysqli_connect("127.0.0.1","","ulchemd"); // Check connection if (mysqli_connect_errno()) { echo "Failed to connect to MySQL: " . mysqli_connect_error(); } $pesquisa =mysqli_real_escape_string($con,$_POST['textquery']); $resposta = mysqli_query($con,"SELECT * from target WHERE molecule.target_text like '%$pesquisa%' "); echo " <b> Search results:"; echo "<table border='5'> <tr> <th>ID</th> <th>target_type</th> <th>name</th> <th>text</th> </tr>"; while($row = mysqli_fetch_array($resposta)) { //este while vai buscar cada linha da tabela num ciclo como um cursor echo "<tr>"; echo "<td>" . $row['tid'] . "</td>"; echo "<td>" . $row['target_type'] . "</td>"; echo "<td>" . $row['name'] . "</td>"; echo "<td>" . $row['target_text'] . "</td>"; echo "</tr>"; } echo "</table>"; mysqli_close($con); ?> 谢谢 :) 解决方法
在您的问题/代码中,index.html和output.php这两个文件中都没有这种类型的错误或异常.
根据错误,您尝试在代码中导入文件.所以请分享您的完整代码. 您也可以尝试更改output.php文件路径,如下所示. xmlhttp.open("POST","http://localhost/projectName/output.php",true); (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |