在jQuery中搜索JSON对象数组中的元素

Question

我有一个JSON对象数组，其中这些对象也可能是JSON对象数组。现在，我需要动态地找到一个值，因为对象的no：不会在数组中相同。例如，

[ 
    { 
    "Plan": { 
     "Node Type": "CTE Scan", 
     "CTE Name": "tab", 
     "Alias": "tab", 
     "Startup Cost": 756.66, 
     "Total Cost": 761.16, 
     "Plan Rows": 67, 
     "Plan Width": 12, 
     "Filter": "(sum > 10000)", 
     "Plans": [ 
     { 
      "Node Type": "Aggregate", 
      "Strategy": "Sorted", 
      "Parent Relationship": "InitPlan", 
      "Subplan Name": "CTE tab", 
      "Startup Cost": 471.70, 
      "Total Cost": 756.66, 
      "Plan Rows": 200, 
      "Plan Width": 8, 
      "Plans": [ 
      { 
       "Node Type": "Merge Join", 
       "Parent Relationship": "Outer", 
       "Join Type": "Inner", 
       "Startup Cost": 471.70, 
       "Total Cost": 684.82, 
       "Plan Rows": 13968, 
       "Plan Width": 8, 
       "Merge Cond": "(p.pr_id = ep.pr_id)", 
       "Plans": [ 
       { 
        "Node Type": "Sort", 
        "Parent Relationship": "Outer", 
        "Startup Cost": 51.37, 
        "Total Cost": 53.17, 
        "Plan Rows": 720, 
        "Plan Width": 4, 
        "Sort Key": ["p.pr_id"], 
        "Plans": [ 
        { 
         "Node Type": "Seq Scan", 
         "Parent Relationship": "Outer", 
         "Relation Name": "product", 
         "Alias": "p", 
         "Startup Cost": 0.00, 
         "Total Cost": 17.20, 
         "Plan Rows": 720, 
         "Plan Width": 4 
        } 
        ] 
       }, 
       { 
        "Node Type": "Sort", 
        "Parent Relationship": "Inner", 
        "Startup Cost": 420.33, 
        "Total Cost": 430.03, 
        "Plan Rows": 3880, 
        "Plan Width": 8, 
        "Sort Key": ["ep.pr_id"], 
        "Plans": [ 
        { 
         "Node Type": "Hash Join", 
         "Parent Relationship": "Outer", 
         "Join Type": "Inner", 
         "Startup Cost": 19.00, 
         "Total Cost": 189.05, 
         "Plan Rows": 3880, 
         "Plan Width": 8, 
         "Hash Cond": "(ep.emp_id = e.emp_id)", 
         "Plans": [ 
         { 
          "Node Type": "Seq Scan", 
          "Parent Relationship": "Outer", 
          "Relation Name": "employee_vs_product", 
          "Alias": "ep", 
          "Startup Cost": 0.00, 
          "Total Cost": 29.40, 
          "Plan Rows": 1940, 
          "Plan Width": 12 
         }, 
         { 
          "Node Type": "Hash", 
          "Parent Relationship": "Inner", 
          "Startup Cost": 14.00, 
          "Total Cost": 14.00, 
          "Plan Rows": 400, 
          "Plan Width": 12, 
          "Plans": [ 
          { 
           "Node Type": "Seq Scan", 
           "Parent Relationship": "Outer", 
           "Relation Name": "employee", 
           "Alias": "e", 
           "Startup Cost": 0.00, 
           "Total Cost": 14.00, 
           "Plan Rows": 400, 
           "Plan Width": 12 
          } 
          ] 
         } 
         ] 
        } 
        ] 
       } 
       ] 
      } 
      ] 
     } 
     ] 
    } 
    } 
] 

在这里，我需要钥匙“Relaion名称”寻找价值......有三个值与关键，我们不知道有没有：存在于它的对象..所以我需要一个解决方案来动态地使用该密钥来查找值。

我试着编码一个递归函数，但我不成功..所以帮助我，如果你知道它。

递归函数我想：

var result1=searchInPlan(jsonarray[0].Plan); 
 
function searchInPlan(node) 
 
{ 
 
\t var result=''; 
 
\t { 
 
\t \t if(node.Plans == undefined) 
 
\t \t { 
 
\t \t \t if(node["Relation Type"] != undefined) 
 
\t \t \t \t result += node["Relation Name"]; 
 
\t \t \t return result; 
 
\t \t } 
 
\t \t else 
 
\t \t { 
 
\t \t \t return result+searchInPlans(node.Plans); 
 
\t \t } 
 
\t } 
 
} 
 

 
function searchInPlans(nodes) 
 
{ 
 
\t console.log("bollo",nodes); 
 
\t var result = ''; 
 
\t for(j=0;j<nodes.length;j++) 
 
\t { 
 
\t \t result += searchInPlan(nodes[j]); 
 
\t } 
 
\t return result; 
 
}

这是工作，但我能够使用该密钥只得到第一个值...我需要所有该键的值。 。帮助我..

Answer 1

这是我的问题的工作代码...

var x=[]; 
 
var y=[]; 
 
var i1=0; 
 
var j1=0; 
 
function scan(obj) { 
 
    var k; 
 
    if (obj instanceof Object) { 
 
     for (k in obj) { 
 
      if (obj.hasOwnProperty(k)) { 
 
       if(k == "Relation Name") 
 
\t \t \t \t { 
 
\t \t \t \t \t x[i1]= obj[k]; 
 
\t \t \t \t \t i1++; 
 
\t \t \t \t \t 
 
\t \t \t \t } 
 
       scan(obj[k]); 
 
      } 
 
     } 
 
    } 
 

 
};