Question

我正在编写一个允许人们将电影添加到数据库中的网站。

目前，用户可以从已经在数据库中的导演中进行选择，也可以创建一个新导演。

            <div class='dropdownrow' id='director_namerow'>
            <div>
                <label for='director_name'>Director:</label>
                <select name='director_name'>
                    <option value='blank' selected>Select...</option>
                        <?php
                        $sql = "SELECT * 
                                FROM director
                                ORDER BY director_name ASC";
                        $director = mysql_query($sql);
                                    while ($directors=mysql_fetch_array($director)) {
                                ?>
                                <option value="<?php echo $directors['director_id']; ?>"><?php echo $directors['director_name']; ?></option>
                                <?php
                                }
                                ?>
                </select>
                <span style = 'color:red;'> *</span>
            </div>
            <div>
                <label for='director_namenew'>Or new director:</label>
                <input type='text' name='director_namenew' size='25' maxlength='128' />
            </div>
        </div>

问题是，如何检查＆＃34; director_namenew＆＃34;并不等于＆＃34; director_name＆＃34;并且director_namenew已经不在数据库中。

此外，是＆＃34; director_namenew＆＃34;不在数据库中，我也需要将它们添加到数据库中。

控制器脚本

  function validateDirector ($formdata) {
   if(($formdata['director_name'] == "blank") && ($formdata['director_namenew'] == "")){
    return false;   
   }
    else if($formdata['hidden_director_name'] == $formdata['director_namenew']){
        echo 'cannot have directors match';
        return false;
    }
    else {
     return true;   
    }
} // TO COMPLETE

谢谢你们， PB。

Answer 1

您可以使用Ajax将前端需求传输到后端，以便分离视图和控制器。

view.html （根据原始代码进行一些修改）：

<script src="checkRepeatName.js"></script>
<div>
  <label for='director_namenew'>Or new director:</label>
  <input type='text' name='director_namenew' size='25' maxlength='128' />
  <span id="notation"></span>
</div>

checkRepeatName.js （用jQuery编写）：

$("input[name='director_namenew']").change(function(){
 $.get("backend.php?dnamenew=" + this.value,function(data,status){
  $(".notation").text(data);
 });
});

backend.php （我使用面向对象的PHP代码样式，并参考@Barmar的推荐）：

<?php
if (isset($_GET['dnamenew'])) {
  $director_namenew = $_GET['dnamenew'];
};

$dbconn = new mysqli(HOST, USERNAME, PASSWORD, DBNAME, PORT);
$sqlStat =  sprintf('SELECT director_id FROM director WHERE director_name = "%s";', $director_namenew);
if ($sqlQuery = $dbconn->query($sqlStat)){
  if ($sqlQuery->num_rows == 0){
    echo 'This is validated database name';
  } else {
    echo 'Invalidated database name!';
  }
}
$dbconn->close;
?>

--------补充一些东西--------
它可能有助于设计动态下拉菜单中的MVC设计模式

view.html ：

<div class='dropdownrow' id='director_namerow'>
  <div>
    <label for='director_name'>Director:</label>
    <select name='director_name'>
      <option value='blank' selected>Select...</option>                                 
    </select>
    <span style = 'color:red;'> *</span>
  </div>
<div>

controllor.js ：

 $("select[name='director_name']").change(function(){
   $.get("optionOutput.php", function(data,status){
     $("select[name='director_name']").append(data); // append the call-back message from back-end file
 });
});

optionOutput.php ：

<?php
include 'dbmng/dbconfig.php';
$dbconn = new mysqli(HOST, USERNAME, PASSWORD, DBNAME, PORT);
$sqlStat = 'SELECT * FROM director ORDER BY director_name ASC;';
if ($sqlQue = $dbconn->query($sqlStat)){
  while($sqlRes = $sqlQue->fetch_assoc()){
    $output = sprintf('<option value = "%d">%s</option>', $sqlResult['director_id'], $sqlRes['director_name']);
    echo $output;
  }
}
$dbconn->close;
?>

Answer 2

回答我自己的问题

if(($formdata['director_name'] == "blank") && ($formdata['director_namenew'] == "")){
            print "<p>Please enter in director information - Use the back button on your browser to rectify this problem.</p>";
            return false;   
           }
            else if($formdata['director_namenew']) {
                $doesntExist = true;
                $db = getDBConnection();
                $directors = $db->query("SELECT director_name FROM director");
                //Check each one
                foreach ($directors as $director){
                    //If the username is already in the DB stop looking
                    if($formdata['director_namenew'] == $director['director_name']){
                        $doesntExist = false;
                        print "<p>Director entered in new director already exists, please enter in new director or select from drop down menu - Use the back button on your browser to rectify this problem.</p>";
                        break;
                    }
                }
                //DB connection closed when PDO object isn't referenced
                //ie setting $db to null closes the connection
                $db = null;
                return $doesntExist;
            }

- 但是，我如何检查下拉和输入字段是否都已填充？

我在想这样的事情：

else if(($formdata['director_name']) && ($formdata['director_namenew'])) {
    print "Please select either new director or director from drop down menu";
}

检查提交以查看是否已提交两个$ _POST变量。

您怎么看？

PHP - 检查<option>值等于<input /> value

2 个答案: