I have the following json file which I'm sending through my mobile application to PHP and on the PHP side I want to Decode it and insert it into a mysql database.
[{"Address":"Somewhere ",
"Area":"Somwhe",
"CreatedBy":null,
"CreatedDate":"\/Date(1419786831365+0530)\/",
"Distance":0,
"EditedDate":"\/Date(-62135596800000+0000)\/",
"Latitude":12.903999947011471,
"Longitude":77.607999974861741,
"Phone1":"80372899",
"Phone2":"993729927",
"Response":null,
"StoreDescriptions":[],
"StoreName":"First"},
{"Address":"Addwmsj",
"Area":"Sbnns",
"CreatedBy":null,
"CreatedDate":"\/Date(1419786863657+0530)\/",
"Distance":0,
"EditedDate":"\/Date(-62135596800000+0000)\/",
"Latitude":12.960867136716843,
"Longitude":77.647689711302519,
"Phone1":"799268299",
"Phone2":"68393973738",
"Response":"Waiting",
"StoreDescriptions":[{"LongNBQuantity":862,
"MeetDate":"\/Date(1419786915048+0530)\/",
"MeetSummary":"Meeting",
"Response":"Negative",
"StoreName":"Ssxond",
"id":1
},
{"LongNBQuantity":8862,
"MeetDate":"\/Date(1419786927673+0530)\/",
"MeetSummary":"Pjsjsbsj",
"Response":"Waiting",
"StoreName":"Ssxond",
"id":2}],
"StoreName":"Ssxond"},
{"Address":"Sumwhere",
"Area":"Righthere",
"CreatedBy":null,
"CreatedDate":"\/Date(1419953186686+0530)\/",
"Distance":0,
"EditedDate":"\/Date(-62135596800000+0000)\/",
"Latitude":12.903999947011471,
"Longitude":77.607999974861741,
"Phone1":"872737288",
"Phone2":"663838828",
"Response":null,
"StoreDescriptions":[],
"StoreName":"NewEntry"}]
As we can see in this I have Array of objects [{},{},{}...{}] and the complexity increases when one of these objects have array of objects like {"abc":"bcd","storedesc":[{},{},{},....{}]} , so this is a little complex when compared to the simple jsons.
My code is not working. Can anybody guide me in the right direction. Thanks
$json = file_get_contents('php://input');
$result = json_decode($json,true);
/*
Database connection setup done here.
*/
foreach ($result as $key => $value) {
if($value){
$sql = "INSERT INTO StoreInfo(name,created_date,edited_date,address,area,ph_num1,ph_num2,response,latitude,longitude) VALUES ($value->StoreName,$value->CreatedDate,'hellyeah',$value->Address,$value->Area,$value->Phone1,$value->Phone2,$value->Response,$value->Longitude,$value->Latitude)";
if($conn->query($sql) === TRUE){
echo "New record inserted";
}
}
}
There are couple of reasons, this not works.
Because you are using the true flag, when you are decoding the JSON, you will get back only arrays, and as I see, you want to use objects in your querys.
In your query do not wrap your wariables with quotes.
So it will be something like this, but please, see my NOTE section!
$sql = "INSERT INTO StoreInfo (name,created_date,edited_date,address,area,
ph_num1,ph_num2,response,latitude,longitude)
VALUES ('".$value["StoreName"]."','".$value["CreatedDate"]."','hellyeah',
'".$value["Address"]."','".$value["Area"]."','".$value["Phone1"]."',
'".$value["Phone2"]."'
,'".$value["Response"]."','".$value["Longitude"]."','".$value["Latitude"]."')";
NOTE
Please escape your variables comes from outside, or use prepared statements to avoid sql injection!
By $result = json_decode($json,true); it is converted to array. true parameter in json_decode() convert JSON to array(). So you should access value by $value['CreatedDate'] instead of $value->CreatedDate and so on.
foreach ($result as $key => $value) {
if($value){
$sql = "INSERT INTO StoreInfo(name,created_date,edited_date,address,area,ph_num1,ph_num2,response,latitude,longitude) VALUES ('".$value['StoreName']."','".$value['CreatedDate']."','hellyeah','".$value['Address']."','".$value['Area']."','".$value['Phone1']."','".$value['Phone2']."','".$value['Response']."','".$value['Longitude']."','".$value['Latitude']."')";
if($conn->query($sql) === TRUE){
echo "New record inserted";
}
}
}
Reference:
json_decode()
Related
I am not able decode and insert the array data in PHP. I tried to functions like decode and foreach, but nothing work.
The $_POST['area'] data:
[area] => [{"text":"DATA1"},{"text":"DATA2"},*,{"text":"DATA3"}]
I am new to PHP, how can I insert these rows of data into my database with pdo's prepared statements?
$area = $_POST['area'];
foreach ($area as $data) {
echo json_decode($data);
}
echo json_decode($area);
However it does not echo.
Ultimately, I want to use the data in this:
$stmt = $con->prepare(INSERT INTO `table` (`areas`) VALUES(:data));
$stmt->execute(array($data));
First things first, you have an invalid json string because of the *,.
$_POST['area'] = '[{"text":"DATA1"},{"text":"DATA2"},*,{"text":"DATA3"}]';
// uh oh, not valid json -----^^
So I'll assume that this is a posting error and run with the following input data:
$_POST['area'] = '[{"text":"DATA1"},{"text":"DATA2"},{"text":"DATA3"}]';
This means that before you can iterate $_POST['area'], you must json_decode() it. (Demo)
An unchanging prepared statement should be declared before entering the loop -- it is designed to be used over and over.
The binding of values to placeholders and the execution of the query is to be done inside the loop.
$stmt = $con->prepare(INSERT INTO `table` (`areas`) VALUES(?));
foreach (json_decode($_POST['area']) as $obj) {
$stmt->execute([$obj->text]);
}
Your json format is incorrect. To do json_decode(); you need correct json format
$post['area'] = [{"text":"DATA1"},{"text":"DATA2"},*,{"text":"DATA3"}];
changed it to
$post['area'] = [{"text":"DATA1"},{"text":"DATA2"},{"text":"DATA3"}];
then do
$ex = json_decode($data, true);
Use This:
$area = json_decode($_POST['area'],true);
You can't echo an array in PHP. You have to use the var_dump() function to do that.
Have a good day ;).
I am not really sure if this is possible, or if there is any alternative way to do this.
The following code takes multiple input from a user and inserts them in different rows, I don't want it like that. I am looking for a way to insert them in the same row.
<?php $test=$_POST['test'];
foreach ($test as $a => $b){?>
<?php echo $test[$a];?>
<?php
if( !$error ) {
$query = "INSERT INTO test_tbl(test) VALUES('$test[$a]')";
$output = mysql_query($query);
if ($output) {
$errTyp = "success";
$errMSG = "Update Posted";
?>
I am aware there is a GROUP_CONCAT function but I can't seem to get it to work in insert statements and from researching I found out it doesn't really work with insert only with select. So is there any way of sorting this mess?
Here is my attempt on the GROUP_CONCAT (Obviously it's a big error that I receive)
$query = "INSERT INTO testing(item) VALUES(GROUP_CONCAT($test[$a]))";
PS I know this is completely against normalization standards but I am doing it since a customer requested it..
If i understood correctly you are inserting in a foreach loop, this will insert each $test[$a] seperately each time the loop runs. ALternatively you can put all $test[$a] values in an array and then insert that array in one row(which is way faster than multiple insert queries).
Here is a way to do it:
$all_test_values = array();
foreach ($test as $a => $b){
$all_test_values[] = $test[$a];
}
$comma_separated = implode(",", $all_test_values);
if( !$error ) {
$query = "INSERT INTO test_tbl(test) VALUES('$comma_separated')";
$output = mysql_query($query);
}
PS: mysql is bad use mysqli instead
I'm trying to handle a POST request from a web service. It's sending an HTTP POST request like this:
{
"latitude":"12.232",
"longitude":"123.323"
}
It's posting to a PHP file on my server. I know that it is hitting the file for sure. However, I'm not getting the data.
In my PHP, I have this (leaving out a bunch of stuff:
$json = file_get_contents('php://input');
$obj = json_decode($json);
$mine ="sixteen"; //using this for a test
$sql = "INSERT INTO rr_emergency (random) VALUES('$obj');";
$result = $dbh->query($sql)->fetchAll(PDO::FETCH_ASSOC);
This makes no change to my database.
If I do this instead:
$sql = "INSERT INTO rr_emergency (random) VALUES('$mine');";
Then "sixteen" is added in the right spot in a new row in my table each time the webservice calls my PHP. This is how I know I'm receiving data.
NOTE: I was trying to simply add $obj into my table just to see the data format that's returned before I tried to properly parse it and put everything where it belongs.
What am I doing wrong here? I think the problem is here ($json = file_get_contents('php://input');), but not sure what else to try.
Thanks.
So there's a few problems
$obj = json_decode($json);
This will return an object. You want an array
$obj = json_decode($json, true);
Then your PDO is incorrect
$sql = "INSERT INTO rr_emergency (random) VALUES(:val);";
$prep = $dbh->prepare($sql);
foreach($obj as $row) $prep->execute([':val' => $row]);
This will insert your data correctly (using a prepared statement) and loop over the JSON return data
You're trying to insert an object, when you really need a string. use:
$obj = json_decode($json, true)
$obj_str = implode(", ", $obj);
$sql = "INSERT INTO rr_emergency (random) VALUES('$obj_str');";
After I posted the above, you added:
I was trying to simply add $obj into my table just to see the data
format
Objects do not inherently convert to strings, so putting $obj within your query doesn't work. The way I store objects in my DB when I've needed to, is to store the JSON notation directly.
$json = file_get_contents("php://input");
$sql = "INSERT INTO rr_emergency (random) VALUES('$json')";
You lose the ability to perform filtering and selecting operations within the object, but it's an effective way to pack away data that you won't need the DB to parse through.
If you need well formatted, easy to read structure:
$obj = json_decode($json);
$obj_str = print_r($obj,true); //store formatted string
$sql = "INSERT INTO rr_emergency (random) VALUES('$obj_str');";
If as you said, all you need to do is "just see the data format", I suggest echoing to the screen or writing to a log file; do one of the following. To print to screen:
print_r($obj);
To write to file:
$filepath = "/path/to/file.txt"
file_put_contents($filepath,print_r($obj,true));
Important note
Entering text directly into your DB queries without escaping it makes you vulnerable to SQL injection attacks. Use prepared statements instead.
I have passed JSON encoded parameters by POST which we have captured and decoded in another PHP file. I have used the following code to do that.
$entityBody = file_get_contents('php://input');
$entityBody = json_decode($entityBody, true);
I have passed the JSON encoded parameters as follows:
{
"id": "5",
"name": "abcd",
"imei": "1234"
}
Actually the number of parameters I am passing by POST may be 15 to 20 which I am going to insert in a table i.e. each of them is a field in the table in mysql database. I am new to JSON and PHP. What method I know is that get value of each parameter after checking whether it is set like following:
if(isset($entityBody['id']))
...
elseif(isset(...))
...
It is clear there will be many if and else when there are many parameters. So is there any way so that I can store the parameters in table in more efficient way. If anyone helps me in doing that I will be really grateful.
use json_decode function to parse the json to an array or object.
$a = json_decode($entityBody);
$a->id;
refer http://php.net/manual/en/function.json-decode.php
If you're using prepared statements you could do something like this. Please note that you should consider how you implement something like this, and change it to suit your needs.
$json = '{"id": "5","name": "abcd","imei": "1234"}';
$array = json_decode($json, true);
if (!count($array) > 0) {
throw new Exception('No params set.');
}
$allowedKeys = array('id','name','imei'); // could be hard coded, or something like: SELECT * FROM yourdb.INFORMATION_SCHEMA.COLUMNS WHERE TABLE_NAME = yourtable
$queryParams = array();
foreach ($array as $key => $value) {
if (in_array($key, $allowedKeys)) {
$queryParams['keys'][] = $key;
$queryParams['values'][] = $value;
}
}
print_r($queryParams);
$query = 'INSERT INTO yourtable (' . implode(', ', $queryParams['keys']) . ') VALUES (?' . str_repeat(',?', count($queryParams['values'])-1) . ')';
print_r($query); // INSERT INTO yourtable (id, name, imei) VALUES (?,?,?)
Then you can just execute it with the $queryParams['values'] as the values :)
There's however no need to insert "id" if you're using auto increment, it will possibly just end up in some strange errors.
Hey guys, i'm currently learning php and I need to do this
$connection = mysql_open();
$likes= array();
foreach($likes as $like)
{
$insert3 = "insert into ProfileInterests " .
"values ('$id', '$like', null)";
$result3 = # mysql_query ($insert3, $connection)
or showerror();
}
mysql_close($connection)
or showerror();
For some reason this does not work =/ I don't know why. $likes is an array which was a user input. I need it to insert into the table it multiple times until all of the things in the array are in.
EDIT I fixed the issue where I was closing it in my foreach loop. mysql_open is my own function btw.
Any ideas?
For one $likes is an empty array in your example, I am assuming you fix that in the code you run.
The second is you close the MySQL connection the first the time the loop would run, which would prevent subsequent MySQL queries from running.
there's no such function as mysql_open
you may need mysql_connect
also $likes variable is empty. so no foreach iterations will execute.
You close the connection within the foreach loop.
Here is the proper formatted code to insert data...You can use this.
// DATABASE CONNECTION
$conn=mysql_connect(HOST,USER,PASS);
$link=mysql_select_db(DATABASE_NAME,$conn);
// function to insert data ..here $tableName is name of table and $valuesArray array of user input
function insertData($tableName,$valuesArray) {
$sqlInsert="";
$sqlValues="";
$arrayKeys = array_keys($valuesArray);
for($i=0;$i < count($arrayKeys);$i++)
{
$sqlInsert .= $arrayKeys[$i].",";
$sqlValues .= '"'.$valuesArray[$arrayKeys[$i]].'",';
}
if($sqlInsert != "")
{
$sqlInsert = substr($sqlInsert,0,strlen($sqlInsert)-1);
$sqlValues = substr($sqlValues,0,strlen($sqlValues)-1);
}
$sSql = "INSERT INTO $tableName ($sqlInsert) VALUES ($sqlValues)";
$inser_general_result=mysql_query($sSql) or die(mysql_error());
$lastID=mysql_insert_id();
$_false="0";
$_true="1";
if(mysql_affected_rows()=='0')
{
return $_false;
}
else
{
return $lastID;
}
}
// End Of Function
While many PHP newbies (myself included) begin working with databases from good ole' mysql_connect/query/etc., I can't help suggest that you look into PDO, PHP Data Objects. Depending on your prior knowledge and programming background, there may be a steeper learning curve. However, it's much more powerful, extensible, etc.; I use PDO in all my production code database wheelings-and-dealings now.