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我正在尝试通过 jQuery 和 AJAX 更新数据库条目。我正在检查我发送的值是否正确 - 但我不确定如何检查数据库未更新的原因。

我的代码如下：

$(document).on("click", ".approve", function(){
    var classes = $(this).parents('div:eq(0)'); // this gets the parent classes.
    i = 0;
    var pros = [];
    classes.find(".prosncons .pros ul li").each(function(){
        pros.push($(this).text());
    });
    var cons = [];
    classes.find(".prosncons .cons ul li").each(function(){
        cons.push($(this).text());
    });
    var notes = classes.find(".notes").text();
    var id = classes.find(".id").text();
    $.ajax({
      type: "POST",
      url: "../scripts/upload.php",
      data: { method: "approve", pros: pros, cons:cons, notes:notes, id:id },
      success: $(this).closest(".approval").remove()
    });
});

PHP::

    require("connection.php");

    $title = $_POST['title'];
    $content = $_POST['content'];
    $pros = $_POST['pros'];
    $cons = $_POST['cons'];
    $notes = $_POST['notes'];
    $method = $_POST['method'];
    $id = $_POST['id'];
    if($method == "approve"){
        $sql = "UPDATE `approval` SET approved = 1 WHERE pros=:pros AND cons=:cons AND notes=:notes AND id=:id";
        $statement = $conn->prepare($sql);
        $statement->execute(array(':pros' => $pros, ':cons' => $cons, ':notes' => $notes, ':id'=> $id));
        echo $conn->error;
    }
    else{
        $sql = "INSERT INTO `approval` VALUES(null, :title, :pros, :cons, :notes, 0)";
        $statement = $conn->prepare($sql);
        $statement->execute(array(':title' => $title, ':pros' => $pros, ':cons' => $cons, ':notes' => $notes));
    }
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1 回答 1

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您不能将数组直接插入 mysql，因为 mysql 不理解 php 数据类型。Mysql 只理解 SQL。因此，要将这个数组插入 mysql 数据库，您必须将其转换为 sql 语句。您可以将您的优点和缺点值更改为逗号分隔的字符串或任何任何类型​​的字符串。

于 2013-09-12T21:28:40.447 回答