php - 为什么HTML表单不将值插入Mysql？

因此，PHP脚本实际上与服务器建立了连接，但是以某种方式无法填充localhost上的数据库。我可能做错了什么？

script.php

<?php
$link = mysqli_connect("127.0.0.1", "root", "", "testdb");

if (!$link) {
echo "Error: Unable to connect to MySQL." . PHP_EOL;
echo "Debugging errno: " . mysqli_connect_errno() . PHP_EOL;
echo "Debugging error: " . mysqli_connect_error() . PHP_EOL;
exit;
}

echo "Success: A proper connection to MySQL was made! The my_db database is
great." . PHP_EOL;
echo "Host information: " . mysqli_get_host_info($link) . PHP_EOL;

$klas = isset($_POST['klasDown']);
$naam = isset($_POST['naamTxt']);

$stmt = $link->prepare("INSERT INTO resultaten (Klas, Naam) VALUES (?, ?");
if (!$stmt) {
echo "false";
} else {
$stmt->bind_param("ss", $klas, $naam);
$stmt->execute();
}


mysqli_close($link);
?>

HTML：

<form id="myForm" method="post" action="script.php">
Naam: <input name="naamTxt" type="text" maxlength="512"
id="naamTxt" class="searchField"/> <br>

Klas: <select name="klasDown">
  <option value="H4A" selected="selected">H4A</option>
  <option value="H4B" >H4B</option>
  <option value="H4C">H4C</option>
  <option value="H4C">H4D</option>
  <option value="H4C">V4A</option>
  <option value="H4C">V4B</option>
  <option value="H4C">V4C</option>
  <option value="H4C">V4D</option>
</select>


<div class="row">
<div class="col-4">1. Kies 1/5 </div>
<div class="col text-center"><input type="radio" name="v1" id="v1_1"
value="Helemaal mee oneens"></div>
<div class="col text-center"><input type="radio" name="v1" id="v1_2"
value="Deels oneens"></div>
<div class="col text-center"><input type="radio" name="v1" id="v1_3"
value="Neutraal"></div>
<div class="col text-center"><input type="radio" name="v1" id="v1_4"
value="Deels mee eens"></div>
<div class="col text-center"><input type="radio" name="v1" id="v1_5"
value="Helemaal mee eens"></div>
</div>

<input type= "submit" style="text-align: center"
onclick="bereken()">
</form>

^以上是HTML表格

Bereken（）函数：

function bereken() {

    var e = 0;
    var c = 0;


    if(document.getElementById('v1_1').checked)  {
      c = c - 0
    }else if(document.getElementById('v1_2').checked)  {
      c += 1;
    }else if(document.getElementById('v1_3').checked)  {
      c += 2;
    }else if(document.getElementById('v1_4').checked)  {
      c += 3;
    }else if(document.getElementById('v1_5').checked){
      c += 4;
    }

    var klas = li.options[li.selectedIndex].value;
    var naam = document.getElementById('searchTxt').value;

    alert(klas + ", " + naam);
}

^这是Javascript bereken（）函数。
显然代码太多，因此必须添加一些文本。

最佳答案

更换

$klas = isset($_POST['klasDown']);
$naam = isset($_POST['naamTxt']);

与

$klas = $_POST['klasDown'];
$naam = $_POST['naamTxt'];

isset将返回一个布尔值，而不是您期望的字符串。

关于php - 为什么HTML表单不将值插入Mysql？，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/45827119/