Javascript 获取jVectormap的值。getElementbyID不工作
我尝试在jVectormap中设置标记。我从数据库中获取内容,并将其写入隐藏的输入字段中。按以下格式返回:Javascript 获取jVectormap的值。getElementbyID不工作,javascript,php,json,getelementbyid,jvectormap,Javascript,Php,Json,Getelementbyid,Jvectormap,我尝试在jVectormap中设置标记。我从数据库中获取内容,并将其写入隐藏的输入字段中。按以下格式返回: {latLng:[52.5200066,13.404954],name:'Berlin'},{latLng:[53.0792962,8.8016937],name:'Bremen'},{latLng:[49.8728253,8.6511929],name:'Darmstadt'},{latLng:[50.1109221,8.6821267],name:'Frankfurt'},{latLn
{latLng:[52.5200066,13.404954],name:'Berlin'},{latLng:[53.0792962,8.8016937],name:'Bremen'},{latLng:[49.8728253,8.6511929],name:'Darmstadt'},{latLng:[50.1109221,8.6821267],name:'Frankfurt'},{latLng:[53.5510846,9.9936818],name:'Hamburg'},{latLng:[54.3232927,10.1227652],name:'Kiel'},{latLng:[50.937531,6.9602786],name:'Köln'},{latLng:[48.30694,14.28583],name:'Linz'},{latLng:[48.1351253,11.5819806],name:'München'},{latLng:[53.6355022,11.4012499],name:'Schwerin'},{latLng:[48.7758459,9.1829321],name:'Stuttgart'},{latLng:[48.0689177,11.6212533],name:'Unterhaching'},,{latLng:[48.2081743,16.3738189],name:'Wien'},
如果我想读这个字段,什么都不会发生。如果我将这一行复制并粘贴到javascript部分,一切都很好
这是有效的:
<script>
$(function(){
var map,
markers = [
{latLng:[52.5200066,13.404954],name:'Berlin'},{latLng:[53.0792962,8.8016937],name:'Bremen'},{latLng:[49.8728253,8.6511929],name:'Darmstadt'},{latLng:[50.1109221,8.6821267],name:'Frankfurt'},{latLng:[53.5510846,9.9936818],name:'Hamburg'},{latLng:[54.3232927,10.1227652],name:'Kiel'},{latLng:[50.937531,6.9602786],name:'Köln'},{latLng:[48.30694,14.28583],name:'Linz'},{latLng:[48.1351253,11.5819806],name:'München'},{latLng:[53.6355022,11.4012499],name:'Schwerin'},{latLng:[48.7758459,9.1829321],name:'Stuttgart'},{latLng:[48.0689177,11.6212533],name:'Unterhaching'},,{latLng:[48.2081743,16.3738189],name:'Wien'},
],
我的问题在哪里(输入中的值是JSON字符串,而不是对象,因此必须将其解析为对象:
$(function(){
var map,
markers = [
JSON.parse(document.getElementById("geodata").value);
],
输入中的值是JSON字符串,而不是对象,因此必须将其解析为对象:
$(function(){
var map,
markers = [
JSON.parse(document.getElementById("geodata").value);
],
所以我改变了所有的事情,现在它几乎开始工作了 我现在使用了“divs”而不是输入。我更改了php函数并返回了正确的JSON Formt(这是我第一次尝试时的错误) 以下是javascript代码:
function GET_PLACES(){
var result_numbers = document.getElementById("result_numbers").value -1;
var text = "";
var result = "";
for (i = 1; i < result_numbers + 2; i++) {
text = JSON.parse(document.getElementById("geodata"+i).innerHTML);
}
return text;
}
console.log(GET_PLACES());
alert(JSON.stringify(GET_PLACES()));
$(function(){
var map,
markers = [GET_PLACES()],
函数GET_PLACES(){
var result\u numbers=document.getElementById(“result\u numbers”).value-1;
var text=“”;
var结果=”;
对于(i=1;ifunction GET_PLACES(){
var result_numbers = document.getElementById("result_numbers").value -1;
var text = "";
var result = [];
for (i = 1; i < result_numbers + 2; i++) {
text = JSON.parse(document.getElementById("geodata"+i).innerHTML);
result.push(text);
}
return result;
}
console.log(GET_PLACES());
alert(JSON.stringify(GET_PLACES()));
$(function(){
var map,
markers = GET_PLACES(),
函数GET_PLACES(){
var result\u numbers=document.getElementById(“result\u numbers”).value-1;
var text=“”;
var结果=[];
对于(i=1;ifunction GET_PLACES(){
var result_numbers = document.getElementById("result_numbers").value -1;
var text = "";
var result = "";
for (i = 1; i < result_numbers + 2; i++) {
text = JSON.parse(document.getElementById("geodata"+i).innerHTML);
}
return text;
}
console.log(GET_PLACES());
alert(JSON.stringify(GET_PLACES()));
$(function(){
var map,
markers = [GET_PLACES()],
函数GET_PLACES(){
var result\u numbers=document.getElementById(“result\u numbers”).value-1;
var text=“”;
var结果=”;
对于(i=1;ifunction GET_PLACES(){
var result_numbers = document.getElementById("result_numbers").value -1;
var text = "";
var result = [];
for (i = 1; i < result_numbers + 2; i++) {
text = JSON.parse(document.getElementById("geodata"+i).innerHTML);
result.push(text);
}
return result;
}
console.log(GET_PLACES());
alert(JSON.stringify(GET_PLACES()));
$(function(){
var map,
markers = GET_PLACES(),
函数GET_PLACES(){
var result\u numbers=document.getElementById(“result\u numbers”).value-1;
var text=“”;
var结果=[];
对于(i=1;iconsole.log(document.getElementById(“geodata”).value);
查看有什么,然后告诉我是的,正确的方式。谢谢,但不幸的是它不起作用。console说“JSON.parse:第1行第3列处应为属性名或“}”。括号中的坐标似乎是问题所在…mhh:/n然后,输入中的值不是格式良好的JSON值。请执行console.log(document.getElementById(“geodata”).value);
查看有什么内容并告诉我