Question

我无法提交表单。我使用类似的代码提交了一个不同的表单，它工作正常，我只是看不出为什么它这次不起作用。

我没有收到任何错误。我尝试过错误报告我没有收到任何错误。表单输入是粘性的，因此页面重新加载，输入仍然存在。

这是我的html表单：

    <form action="evenement_maken.php" method="POST" enctype="multipart/form-data">

                <input type="text" name="ev_naam" class="input-lg form-control" value='<?php echo (isset($_POST['ev_naam']) ? $_POST['ev_naam'] : "" );?>'> 

             <input type="text" name="ev_datum">

                <input type="text" name="ev_adres" class="input-lg form-control" placeholder="Vul hier het adres van het evenement in..." value='<?php echo (isset($_POST['ev_adres']) ? $_POST['ev_adres'] : "" );?>'>

                <textarea class="input-lg form-control" rows="10" name="ev_omschrijving" id="textarea" placeholder="Korte omschrijving van het evenement...">
          <?php  
              if(isset($_POST['ev_omschrijving'])){
              echo htmlentities($_POST['ev_omschrijving'], ENT_QUOTES);
              }
          ?>
        </textarea>
            <button type="submit" class="pull-right btn btn-danger" name="submit">Opslaan</button>   
        </form>

我的php代码：

<?php

 $ev_naam = $ev_datum = $ev_omschrijving = $ev_adres = "";

if(isset($_POST['submit'])) {
    $ev_naam = mysqli_real_escape_string($conn, $_POST['ev_naam']);
    $ev_datum = mysqli_real_escape_string($conn, $_POST['ev_datum']);
    $ev_omschrijving = mysqli_real_escape_string($conn, $_POST['ev_omschrijving']);
    $ev_adres = mysqli_real_escape_string($conn, $_POST['ev_adres']);


    if ($ev_naam=='') {
      echo "<script>alert('Vul alsjeblieft alle velden in!')</script>";
      exit();// zorgt ervoor dat de rest van het script niet wordt uitgevoerd

    } else {

      $insert_evenementen = "INSERT INTO evenementen (ev_naam,ev_datum,ev_omschrijving,ev_adres)
              VALUES ( '$ev_naam','$ev_datum','$ev_omschrijving','$ev_adres')";

        $run_evenementen = mysqli_query($conn, $insert_evenementen);

        if (mysqli_query($conn, $insert_evenementen)) {
          echo "<script>alert('Post is succesvol opgeslagen!')</script>";
          echo "<script>window.open('evenement_maken.php','_self')</script>";
        }
    }

}
?>

这是正确提交的表单（仅将img上传到ftp不起作用）：

    <form action="post_maken.php" method="post" enctype="multipart/form-data">
        <h4>Titel: </h4>
            <input type="text" name="post_titel" class="input-lg form-control" value='<?php echo (isset($_POST['post_titel']) ? $_POST['post_titel'] : "" );?>' required> 
        <h4>Inhoud: </h4>
            <textarea class="input-lg form-control" rows="10" name="post_inhoud" id="textarea" required>

          if(isset($_POST['post_inhoud'])){
          echo htmlentities($_POST['post_inhoud'], ENT_QUOTES);
          }
      ?>
    </textarea>
        <h4>Categorie:</h4>
            <select class="form-control" name="categorie_id" >
        <option value="null" >selecteer een categorie...</option>
        <?php

                $categorie = mysqli_query($conn, "SELECT * FROM categorie");

                while ($cat_row=mysqli_fetch_array($categorie, MYSQLI_ASSOC)) {

                    $cat_naam=$cat_row['cat_naam'];

                  echo "<option value='$cat_naam'>$cat_naam</option>";

                } 
    </select>

        <h4>Afbeelding toevoegen</h4>
            <div class="input-group">
                <span class="input-group-btn">
                    <span class="btn btn-primary btn-file">
                        Zoeken&hellip; 
                    </span>
                </span>

            </div>

      <input type="file" name="post_img"/>


            <p class="help-block">Voeg een afbeelding voor je blogpost toe.</p>
            <br>
        <button type="submit" class="pull-right btn btn-danger" name="submit">Opslaan</button>   
    </form>

和php代码：

<?php 

$post_titel = $post_datum = $post_inhoud = $categorie_id = "";

if(isset($_POST['submit'])) {

$post_titel = mysqli_real_escape_string($conn, $_POST['post_titel']);
$post_datum = mysqli_real_escape_string($conn, date('m-d-y'));
$post_inhoud = mysqli_real_escape_string($conn, $_POST['post_inhoud']);
$categorie_id = mysqli_real_escape_string($conn, $_POST['categorie_id']);
$post_img = mysqli_real_escape_string($conn, $_FILES['post_img']['name']);
$post_img_tmp = mysqli_real_escape_string($conn, $_FILES['post_img']['tmp_name']);

if ($post_titel=='' || $categorie_id=='null' || $post_inhoud=='') {

  echo "<script>alert('Vul alsjeblieft alle velden in!')</script>";
  exit();

} else {

  move_uploaded_file($post_img_tmp, "post_img/$post_img");

  $post_bron = 0;

  $post_datum = date("y-m-d"); 

  $insert_posts = "INSERT INTO post (post_title,post_inhoud,post_datum,categorie_id, post_img, post_bron)
          VALUES ( '$post_titel','$post_inhoud','$post_datum','$categorie_id','$post_img','$post_bron')";

    $run_posts = mysqli_query($conn, $insert_posts);

    if (mysqli_query($conn, $insert_posts)) {
      echo "<script>alert('Post is succesvol opgeslagen!')</script>";
      echo "<script>window.open('post_maken.php','_self')</script>";
    }
}

}
?>

我正从两个数据库（连接工作）中检索值并在网站上显示它，这也有效。我正在使用bootstrap 3.

我的数据库表的屏幕截图： database screenshot

谁能看到我做错了什么？我一直盯着这几个小时。

Answer 1

您可能有SQL错误。当它们发生时检查并输出mysqli_erros总是很好的。切换你的代码来做到这一点

if (mysqli_query($conn, $insert_evenementen)) { 
    // query was succesful
}else{ 
    echo mysqli_erro($cnon); // sthing went wrong
}

您的代码对SQL注入是开放的，我建议您查看准备好的语句

<?php
//procedural example from http://php.net/manual/en/mysqli.prepare.php
$city = "Amersfoort";

/* create a prepared statement */
if ($stmt = mysqli_prepare($link, "SELECT District FROM City WHERE Name=?")) {

    /* bind parameters for markers */
    mysqli_stmt_bind_param($stmt, "s", $city);

    /* execute query */
    mysqli_stmt_execute($stmt);

    /* bind result variables */
    mysqli_stmt_bind_result($stmt, $district);

    /* fetch value */
    mysqli_stmt_fetch($stmt);

    printf("%s is in district %s\n", $city, $district);

    /* close statement */
    mysqli_stmt_close($stmt);
}

php表单没有提交，页面只是重新加载

1 个答案: