我需要運行 PHP function 並在進行下拉選擇並單擊按鈕時顯示數據

Question

我正在嘗試調用 function 並使用 PHP 顯示數據。 這就是我希望我的代碼工作的方式。

用戶從下拉列表中選擇 state，單擊提交按鈕。 點擊 function getState 被調用並運行

SELECT * FROM table_name WHERE state LIKE '$state'

'$state' is a variable based off of the state selected in the Dropdown.

Once A user selects a state from the drop down and clicks the submit button the page should run the PHP function and display the state (selected) data in a table

基本上，如果用戶從下拉列表中選擇“Indiana”並單擊提交按鈕

查詢將運行“SELECT * FROM table_name WHERE state LIKE 'Indiana'”；

並顯示一個表格，其中 Header 為“州”，數據為“印第安納州”

關於如何 go 對此有何建議？ 我也願意使用 javaScript 或 jQuery 來調用 function。 另外，我不確定我的語法是否正確。

    <label for="states">Choose State</label>

    <select name="states" id="stateSelect">
      <option value="Kentucky">New York</option>
      <option value="Indiana">Indiana</option>
      <option value="Ohio">Ohio</option>
      <option value="South_Carolina">South Carolina</option>
      <option value="North_Carolina">North Carolina</option>
    </select>
    <br>
    <form method="post">
      <input type="submit" name="getState" value="getState" />
    </form>

<?php


    function getState($state){
      $c = mysqli_connect("localhost", "root", "test", "database_name");


    // Check connection
    if($c === false){
    die("ERROR: Could not connect. " . mysqli_connect_error());
    }
      $sql = "SELECT * FROM table_name WHERE state LIKE '$state'";
      $result = mysqli_query($c, $sql);
      echo "<table>";
      echo "<tr>";
      echo "<th>State</th>";

      while ($rows = mysqli_fetch_array($result, MYSQLI_ASSOC)){
        echo "<tr>";
        echo "<td>" . $row['state'] . "</td>";
        echo "</tr>";
        }
        echo "</table>";
    }

    // Close connection
    mysqli_close($c);

      ?> 

Answer 1

試試這個，我只是改進你的代碼跳，如果你的數據庫沒問題，這將對你有所幫助

 <form action="thispage.php" method="post">// in action write your file name where you use this code
  <label for="states">Choose State</label>

  <select name="states" id="stateSelect">
    <option value="Kentucky">New York</option>
    <option value="Indiana">Indiana</option>
    <option value="Ohio">Ohio</option>
    <option value="South_Carolina">South Carolina</option>
    <option value="North_Carolina">North Carolina</option>
  </select>
  <br>
  <input type="submit" name="getState" value="getState" />
</form>

<?php

  if(isset($_POST['getState'])){
    $state = $_POST['states'];
    getState($state);
  }


  function getState($state)
  {
    $c = mysqli_connect("localhost", "root", "test", "database_name");

    // Check connection
    if($c === false){
      die("ERROR: Could not connect. " . mysqli_connect_error());
    }
    $sql = "SELECT * FROM table_name WHERE state LIKE '$state'";
    $result = mysqli_query($c, $sql);
    echo "<table>";
    echo "<tr>";
    echo "<th>State</th>";

    while ($rows = mysqli_fetch_array($result, MYSQLI_ASSOC)){
      echo "<tr>";
      echo "<td>" . $row['state'] . "</td>";
      echo "</tr>";
    }
    echo "</table>";
  }

  // Close connection
  mysqli_close($c);
?>