How can I display user search results onto the same page inside a table?
Question
I'm trying to get some user search result data to print into a table on the same page the search was made from. I've searched for solutions here, but all of the examples are using MySql. I'm currently just searching from a single .log file in the project. I've used an AJAX call to start the search from a button click. I can see that the search input does return a match from the .log file.
I'm fairly new to PHP (2 weeks!) and would love any suggestions on how to achieve this.
error-lookup.php
<?php
require_once $RootPath . "p/check-session.inc";
$RootPath = "";
$User = "";
$RetCode = 200;
$sProcess = "error-lookup.php";
?>
<!doctype html>
<html lang="en">
<head>
<title>MyTools</title>
<?php include_once ($RootPath."p/html-head.inc"); ?>
</head>
<script>
function verifyAccount() {
var pageData = {};
errorData = $('#userInput').val();
pageData['errorData'] = errorData;
var param = JSON.stringify(pageData);
$.ajax({
global: false,
"url": "p/search-log.php",
data: pageData,
cache: false,
"type": "POST",
dataType: "json",
"complete": function (selectionData) {
console.log(selectionData);
try {
var resultData = $.parseJSON(selectionData.responseText);
var output = "";
for (var i = 0; i < resultData.length; i++) {
var output = "<td>" + resultData[i].Id + "</td><br>" + "<td>" + resultData[i].Date + "</td><br>" + "<td>" + resultData[i].User + "</td><br>" + "<td>" + resultData[i].Message + "</td><br>" + "<td>" + resultData[i].StackTrace + "</td>";
$("dataTable").append(output);
}
document.write(output);
$("dataTable").html(resultData);
} catch (err) {
showError(1,err);
}
}
});
}
</script>
<body>
<div class="page-header">
<div class="container">
<div class="row-fluid">
<medium>Platform error code lookup</medium>
</div>
</div>
</div>
<form class="container">
<input type="radio" name="tags[]" value="East-Coast" class="east-radio"/>
<label >East Coast Servers</label><br/>
<input type="radio" name="tags[]" value="West-Coast" class="west-radio"/>
<label >West Coast Servers</label>
</form><br>
<div class="input-group mb-3">
<form class="container" action="search-log.php" method="post">
<input class="form-control" id="userInput" type="text" name="input_tag" placeholder="Enter error code or user name">
<div class="input-group-append">
<input class="btn btn-outline-secondary" value="Submit" name="submit_button" type="button" onclick="verifyAccount()"></input>
</div>
</form>
<table class="table" id="dataTable">
<tbody>
<tr>
<th>Date</th>
<th>Error Id</th>
<th>User</th>
<th>Message</th>
<th>StackTrace</th>
</tr>
<td></td>
<td></td>
<td></td>
<td></td>
<td></td>
</tbody>
</table>
<div class="container">
<div id="test"></div>
</div>
</body>
</html>
search-log
<?php
$errorData = $_REQUEST["errorData"];
$errors = [];
$lines = file("../MyWeb.log", FILE_IGNORE_NEW_LINES);
foreach($lines as $line) {
$tmp = [];
$tmp["Date"] = substr($line, 0, 19);
$pos = strpos($line, '{');
$jsonObject = json_decode(substr($line, $pos),true);
$tmp["Id"] = $jsonObject["Id"];
$tmp["User"] = $jsonObject["User"];
$tmp["Message"] = $jsonObject["Message"];
$tmp["ServerName"] = $jsonObject["ServerName"];
$tmp["StackTrace"] = $jsonObject["StackTrace"];
array_push($errors, $tmp);
}
$results=[];
foreach($errors as $error){
if (strpos($error["Date"], $errorData) > -1) {
array_push($results, $error);
continue;
}
if (strpos($error["Id"], $errorData) > -1) {
array_push($results, $error);
continue;
}
if (strpos($error["User"], $errorData) > -1) {
array_push($results, $error);
continue;
}
if (strpos($error["Message"], $errorData) > -1) {
array_push($results, $error);
continue;
}
if (strpos($error["ServerName"], $errorData) > -1) {
array_push($results ,$error);
continue;
}
if (strpos($error["StackTrace"], $errorData) > -1) {
array_push($results, $error);
continue;
}
}
print_r($results);
2 answers
solution1
0 2018-07-05 14:28:50
Check if you are getting json in valid format before parsing it.
$.ajax({
global: false,
"url": "p/search-log.php",
data: pageData,
cache: false,
"type": "POST",
"complete": function (selectionData) {
// check if you are getting valid JSON before parsing it
console.log(selectionData)
try {
var resultData = $.parseJSON(selectionData.responseText);
$("#test").html(resultData);
} catch (err) {
showError(1,err);
}
}
});
send data in json format
// send json from your php script
echo json_encode($results);
NOTE: you can use right click -> inspect -> network tab to check what your server is returning and whether the ajax call is being made.
Hope it helps!
solution2
0 ACCPTED 2018-07-06 14:49:18
The flow is as follows:
-> $.ajax send a post with parameters;
-> You receive at your PHP script like any other post, does not need to json decode;
-> You select rows or anything you want to do, make a array or object structure, then return it with json_encode and Content-Type: application/json;
-> At success function, response will become a JS object as long as dataType is defined to Json;
Ok. First, to convert to JSON, you need to make an array or object, then, return it with json_encode. It will convert it into a JSON valid string that jQuery will parse correctly. Let's say you have a structure like: array ('name' => 'my_name'); When you JSON encode it, it will become a JSON string like { "name": "my_name" }
At our complete or success function, you will have responseText.name where the value will be "my_name". Now with it, you can do anything, like create a domNode with a div and insert it.
Eg: you could create a table with id, then:
$('#id-of-my-table').append('<tr><td>' + rensponseText.name + '</td></tr>');
Of course this is a raw example, just to illustrate a really simple usage.
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