Question

我是新来的。所有在线示例都无法帮助我调试。我想念什么？这个想法是用户输入他们想要转换为密码的内容，然后js函数为他们加密。 js本身运行良好。尝试使函数Crypt（）与html一起使用时，断开连接会发生。

override func updateConstraints() {
    /// Check if views already have constraints
    if !viewsHaveConstraints {
        viewsHaveConstraints = true


        /* 
        Here I set the hugging and compression priority for the 3 views in the UIStackView
        */


        let margins = safeAreaLayoutGuide

        /*  Set vstackView constraints */

        /// vStackView width anchor equal to width of margin less constant
        vStackView.widthAnchor.constraint(equalTo: margins.widthAnchor, constant: -16.0).isActive = true

        /// vStackView leading anchor constraint
        vStackView.leadingAnchor.constraint(equalTo: margins.leadingAnchor, constant: 8.0).isActive = true

        /// vStackView trailing anchor constraint
        vStackView.trailingAnchor.constraint(equalTo: margins.trailingAnchor, constant: 8.0).isActive = true

        vStackView.heightAnchor.constraint(equalTo: margins.heightAnchor).isActive = true
    }
    super.updateConstraints()
}

Answer 1

您已将按钮放置在<head>标签内。其他一切都很好。将其放在<body>内就可以了！

function Crypt() {
  var input = document.getElementById('input1').value;
  var resultArray = [];
  for (var i = 0; i < input.length; i++) {
    if (input[i] === 'a') {
      resultArray.push('p');
    } else if ("input" [i] === 'b') {
      resultArray.push('l');
    } else if (input[i] === 'c') {
      resultArray.push('m');
    } else if (input[i] === 'd') {
      resultArray.push('n');
    } else if (input[i] === 'e') {
      resultArray.push('k');
    } else if (input[i] === 'f') {
      resultArray.push('o');
    } else if (input[i] === 'g') {
      resultArray.push('i');
    } else if (input[i] === 'h') {
      resultArray.push('j');
    } else if (input[i] === 'i') {
      resultArray.push('t');
    } else {
      resultArray.push(' ');
    }
  }
  document.write(resultArray.join(''));
}

<input type="text" id="input1"/>
<button onclick="Crypt()">Submit</button>

使用用户输入功能时出现问题

1 个答案: