2014-12-03 111 views
-3

我想從我的數據庫檢索數據使用和HTML表單和PHP連接到我的sql-workbench。我有一個成功的數據庫連接,我的檢索發佈在我的網頁頂部。我不知道如何讓它在提交表單函數上運行sql查詢。 任何幫助,將不勝感激,謝謝。使用HTML表單來搜索和從mysql工作臺檢索數據由php

<!DOCTYPE html> 
 
<html> 
 
    
 
    <head data-gwd-animation-mode="quickMode"> 
 
    <title>Test_webpage</title> 
 
    <meta http-equiv="Content-Type" content="text/html; charset=utf-8"> 
 
    <meta name="generator" content="Google Web Designer 1.1.3.1119"> 
 
    
 
    </head> 
 
    
 
    <body> 
 
    
 
    
 
<?php 
 

 

 
$servername = "localhost"; 
 
$username = "root"; 
 
$password = "006"; 
 
$dbname = "mydb"; 
 

 
// Create connection 
 
$con = new mysqli($servername, $username, $password, $dbname); 
 
    
 
     //$con->close(); 
 
     //$query = "SELECT * FROM US_Cars"; 
 

 
if ($con->connect_error) { 
 
    die("Connection failed: " . $con->connect_error); 
 
} 
 

 
$sql = "SELECT Manufacturer,VIN,Color,Model,Fuel_Type,State_of_Origin FROM mydb.US_Cars;"; 
 
$result = $con->query($sql); 
 

 

 
if ($result->num_rows > 0) { 
 
    // output data of each row 
 
    while($row = $result->fetch_assoc()) { 
 
     echo "Manufacturer: " . $row["Manufacturer"]. " VIN: " . $row["VIN"]. " Color:" . $row["Color"]. " Model:" . $row["Model"]. " Fuel Type:" . $row["Fuel_Type"]. " State Location:" . $row["State_of_Origin"]. "<br>"; 
 
    } 
 
} else { 
 
    echo "0 results"; 
 
} 
 

 
?> 
 

 

 

 

 
    <div id="first"> 
 
     <p>Enter Criteria Below to Search.</p> 
 
    </div> 
 
    <br> 
 
    <div id="second"> 
 
     <p class="gwd-p-j2b6">Select a Manufacturer</p> 
 

 

 
<!-- start of form --> 
 
     <form class="gwd-form-yuwo" method="post" action="<?search2.php ?>"> 
 
     <select class="auto-style3 gwd-select-u9g0"> 
 
      <option value="GM">GM</option> 
 
      <option value="Ford">Ford</option> 
 
      <option value="Chrysler">Chrysler</option> 
 
      <option value="Tesla">Tesla</option> 
 
     </select>appreaciated 
 
     <br> 
 
     <br> 
 
     <p class="gwd-p-cy97">Select a Dealership Location</p> 
 
     <select class="auto-style3 gwd-select-dkho"> 
 
     <option value="">Virginia</option> 
 
     <option value="">Maryland</option> 
 
     <option value="">New Jesery</option> 
 
     <option value="">Kentucky</option> 
 
     </select> 
 
     
 
     <br> 
 
     <input type="submit" value="Submit" class="auto-style2"> 
 
     </form> 
 
    </div> 
 
    <div id="third"> 
 
     <img id="pic1" src="page2_1.jpg" style="width:500px;" height="236px" > 
 
    </div> 
 
    <div id="forth"> 
 
     <img src="tesla.png" style="width:125px;height:250px"> 
 
    </div> 
 
    <div id="fifth"> 
 
     <footer> 
 
     <p> 
 
      <a href="">U.S. Car data Entry</a> 
 
     </p> 
 
     <p> 
 
      <a href="">About Page</a> 
 
     </p> 
 
     </footer> 
 
    </div> 
 
    </body> 
 

 
</html>

+0

首先,'action =「<?search2.php?>」'這是值得懷疑的。如果沒有設置短標籤,這個'action =「<?php search2.php?>」'可以被修改。如果你想從數據庫中檢索,你需要分配POST變量並命名你的表單元素。 – 2014-12-03 02:39:04

回答

0
<form class="gwd-form-yuwo" method="post" action="<?search2.php ?>"> 

是你的問題..敢肯定,被認爲是

<form class="gwd-form-yuwo" method="post" action="search2.php"> 

這將送你按下提交按鈕時search2.php。 ..然後在search2.php中,您將從$ _POST獲取表單值,然後運行您的查詢。

這就是說,我認爲這不是你想要做的,你真正想要做的是更新網頁內容的列表,而無需鏈接。即在按鈕上按下更新頁面頂部的html以使用來自表格的prams列出查詢結果。如果這是你正在嘗試做的事情,那麼你需要查找ajax ...你的提交按鈕不會有直接的動作,而是運行一個javascript函數,它會發送一個ajax請求到search2.php,查詢,然後你的Java腳本會更新響應中的innerhtml。