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Javascript 获取json无“访问控制允许来源”错误-电话间隔_Javascript_Php_Jquery_Json_Cordova - Fatal编程技术网

Javascript 获取json无“访问控制允许来源”错误-电话间隔

Javascript 获取json无“访问控制允许来源”错误-电话间隔,javascript,php,jquery,json,cordova,Javascript,Php,Jquery,Json,Cordova,在尝试检索和显示phonegap应用程序中的信息时,我不断遇到此错误 这是我得到的错误: 获取send@jquery-2.2.0.js:9172jQuery.extend.ajax@jquery-2.2.0.js:8653jQuery.anonymous function@jquery-2.2.0.js:8804jQuery.extend.getJSON@jquery-2.2.0.js:8785匿名function@bookmarks.html:44 bookmarks.html:1无法加载XM

在尝试检索和显示phonegap应用程序中的信息时,我不断遇到此错误

这是我得到的错误:

获取send@jquery-2.2.0.js:9172jQuery.extend.ajax@jquery-2.2.0.js:8653jQuery.anonymous function@jquery-2.2.0.js:8804jQuery.extend.getJSON@jquery-2.2.0.js:8785匿名function@bookmarks.html:44 bookmarks.html:1无法加载XMLHttpRequest。请求的资源上不存在“Access Control Allow Origin”标头。因此,不允许访问源“null”。响应的HTTP状态代码为404

代码如下:

<script>
//populates product container
$(" #productContainer").html('');
$.getJSON("http://192.168.0.4:3000/Application/www/retrieveSymbol.php", function(data){  //retrieves json array
  $.each(data, function(i, field){          //loops through array
    $("#productContainer").append(          //creates product box filling it with data
      "<div id='productBox'>" + "<div id='name'>" + field[1] + " </div>" + "<br>" +
      "<div id='image'>" + "<img id='image' src='Images/" + field[5] + "'/>" + "</div>" +
      "<div id='description'>" + "<b>Description:</b> "  + "<br>" + field[2]  + " </div>" + "" +
      "<div id='price'>" + "£" + field[3] + " </div>" + "<br><br>" +  "<br>" +

      //gets name and price and sends data to basket when button is clicked
      "<form action='Basket.php' method='post' id='add'>" +
      "<input type='hidden' name='itemName' value=" + field[1] + "/>" +
      "<input type='hidden' name='itemPrice' value=" + field[3] + ">" +
      "<input id='addButton'type='submit' value='Add to Cart'>" +
      "</form>" + "</div>"

    );
  });
});
</script>
receivesymbol.php

<?php

//connect to the database
$mysqli = NEW MySQLi ('localhost','root','','contentdatabase');
//query database
$resultSet = $mysqli->query("SELECT * FROM items");
//count the rows
if($resultSet->num_rows != 0) {
  //turn the results into an array
  $rows = $resultSet->fetch_all();
  echo json_encode($rows);

}else{
  echo "{no connection}";
}
?>

有人能指出发生这种情况的原因吗?

这是大多数PhoneGap项目中常见的典型跨源错误。请查看此答案以了解更多详细信息和解决方案


使用此chrome扩展,您可以从任何来源请求任何带有ajax的站点

接收Symbol.php添加Access Control Allow Origin标头-这似乎就是它抱怨的服务器脚本中的可启用CORS。请参阅是否在设备或模拟器中进行测试?另外,请在chrome windows中尝试此操作-