Question

我试图获得一个发送并收到ajax编码，但我得到一个错误，preco不是Complista2页面中的索引。这是代码：

起始页：

$(function() {      
        $("#b1").click(function() {
            $.ajax({
                type: "POST",
                url: "complista2.php",
                data: $("#preco").val(),                
                success: $.get("complista2.php", function(data) {
                    $("#paramount").html(data);
                })
            });
        }); 
    });

<input id="preco" type="text" name="preco" value='250 AND 500'><br><br>

这是Complista2页面：

<?php

$username = "root";
$password = "";
$server = "localhost";
$database = "VRealidade";
$conn = mysqli_connect($server, $username, $password, $database);
if (!$conn) {
    die("Erro: " . mysqli_connect_error());
} else {    
    }

$preco = $_GET['preco'];
$sql = "SELECT * FROM computadores WHERE preco BETWEEN $preco"; 
$result = mysqli_query($conn, $sql);
echo mysqli_error($conn);
$row = mysqli_fetch_assoc($result);
echo "<div class='entrada'>$row[nome]</div>"; 
while($row = mysqli_fetch_assoc($result)) {
    echo "<div class='entrada'>$row[nome]</div>";
}       

?>

编辑：这是表单，但在此代码中没有任何含义。

<form id="pesqmenu2" method='GET'>
    <input id="preco" type="text" name="preco" value='250 AND 500'>
</form>

Answer 1

<?php

$username = "root";
$password = "";
$server = "localhost";
$database = "VRealidade";
$html = "";

//check if preco initialized, and filter for default and special chars as SQL injection ...
if(isset($_POST['preco']) && filter_input(INPUT_POST, 'preco', FILTER_DEFAULT , FILTER_SANITIZE_SPECIAL_CHARS)) {

  try {
    $preco = $_POST['preco'];
    $conn = mysqli_connect($server, $username, $password, $database);

    if (mysqli_error()) {
      $error = "MySQLi error " . mysqli_error() . ": " . mysqli_error() . "\n<br>When executing:<br>\n$query\n<br>";
      exit();
    }


    $sql = "SELECT * FROM computadores WHERE preco BETWEEN $preco";
    $result = mysqli_query($conn, $sql);
    $row = mysqli_fetch_assoc($result);

    $html .= "<div class='entrada'>$row[nome]</div>";

    while ($row = mysqli_fetch_assoc($result)) {
      $html .= "<div class='entrada'>$row[nome]</div>";
    }

    //ouput html
    header('Content-Type: application/json');
    echo json_encode($html);

  } catch (Exception $e) {
    echo 'Exception reçue : ', $e->getMessage(), "\n";
  }
}

您需要更改表单，问题是不要发送到页面.php

<form id="pesqmenu2" name="pesqmenu2" method="post">
  <input id="preco" type="text" name="preco" value='250 AND 500'>
  <input type="submit" id="b1" name="b1" value="submit"/>
</form>

Ajax改变了：

$(function() {   
    $.ajax({
        method: "POST",
        url: "complista2.php",
        data: { preco: $("#preco").val() }
    })
    .done(function( data ) {
        $("#paramount").html(data);
    });
});

Answer 2

我找到了解决方案。我所要做的就是在点击事件中添加这个：

$.post("Complista2.php/?preco="+$("#preco").val(), function(data) {
        $("#paramount").html(data);
});

ajax - 发送和接收数据错误

2 个答案: