Answers

• allan Posts: 65,287Questions: 1Answers: 10,824 Site admin

  September 2024

  Since 2.0 the ajax.dataSrc option can be given as an object to allow it to work with data, draw and the other properties. You need to specify a function for each one.

  The alternative, and it might be best in this case, is to specify ajax as a function - make your own Ajax call, decode the result and then pass it to the callback function.

  Allan
• sabareesh_pix Posts: 3Questions: 1Answers: 0

  September 2024

  I was able to resolved the issue by doing:

  dataSrc: function(json) {
      const decrypted = decryptData(json);
      if (decrypted != null && decrypted != "") {
          const decryptedData = JSON.parse(decrypted);
          json.recordsTotal = decryptedData.recordsTotal || 0
          json.recordsFiltered = decryptedData.recordsFiltered || 0
          json.data = decryptedData.data || [];
      }
      return json.data;
  }
  

  But, what I don't understand is that.. It is not working when I replace the whole response with decrypted data as below. I have no any additional keys with my response. Only data, recordsTotal, recordsFiltered. I like to know what is the reason though.

  dataSrc: function(json) {
      const decrypted = decryptData(json);
      if (decrypted != null && decrypted != "") {
          const decryptedData = JSON.parse(decrypted);
          json.recordsTotal = decryptedData.recordsTotal || 0
          json.recordsFiltered = decryptedData.recordsFiltered || 0
          json.data = decryptedData.data || [];
      }
      return json.data;
  }
  
• allan Posts: 65,287Questions: 1Answers: 10,824 Site admin

  September 2024

  I don't actually see any difference between the two code blocks? Most likely though the issue was that dataSrc expects an array on return, not the object with recordsTotal etc.

  Allan
• sabareesh_pix Posts: 3Questions: 1Answers: 0

  September 2024

  Apologies, I mistakenly pasted the same code samples above, so please disregard them.

  When I set dataSrc as shown below, it doesn't include recordsTotal and recordsFiltered:

  dataSrc: function(json) {
      const decrypted = decryptData(json);
      if (decrypted != null && decrypted != "") {
          json = JSON.parse(decrypted);
      }
      return json.data;
  }
  

  However, if I don't completely override the JSON object but instead replace only specific properties, like recordsTotal and recordsFiltered, it works fine. Here’s an example where I override only these properties without fully replacing the JSON:

  dataSrc: function(json) {
      const decrypted = decryptData(json);
      if (decrypted != null && decrypted != "") {
          const decryptedData = JSON.parse(decrypted);
          json.recordsTotal = decryptedData.recordsTotal || 0
          json.recordsFiltered = decryptedData.recordsFiltered || 0
          json.data = decryptedData.data || [];
      }
      return json.data;
  }
  
• allan Posts: 65,287Questions: 1Answers: 10,824 Site admin

  September 2024

  json = JSON.parse(decrypted);
  

  This assigned a different object to the local json variable. It does not effect the object that had been passed in and assigned to the local json variable.

  The reason why json.recordsTotal = decryptedData.recordsTotal || 0 works, is that you are mutating the original object, not just a local variable pointer.

  Allan