I am trying to print a python variable to a php page. I have my php script status.php shown below
<?php
$result = json_decode(exec('python status.py'), true);
echo $result;
?>
This php page calls the status.py python page which is listed below
import json
import RPi.GPIO as GPIO
import time
GPIO.setmode(GPIO.BCM)
GPIO.setuip(23, GPIO.IN, pull_up_down=GPIO.PUD_UP)
input_state = GPIO.input(23)
if input_state == False:
print("Door is closed")
D = 1
elif input_state == True:
print("Door is open")
D = 0
print json.dumps(D)
I can get the shell command to print the php variable which when the door is closed comes out to be
Door is closed
1
However I can't get the variable 1 to print to the php page when I run it from the browser. Any help would be greatly appreciated!
By changing
$result = json_decode(exec('python status.py'), true);
to
$result = json_decode(exec('sudo python status.py'), true); it enabled the variable to be passed from the python script to the php script. It seems there was an access problem when accessing the file via webrowser.
Related
please help me... I have no idea.
This is part of my php code.
$result;
$resultD;
$arg= $ord." ".$time2_int;
$result=shell_exec('python3 decemberlist.py '.$arg);
$resultD=json_decode($result,true);
echo $resultD['song'];
And This is part of my python code.
import sys
import json
msg=''
def songName(a, b):
global msg
//
return msg
ret=songName(sys.argv[1], sys.argv[2])
rett={'song':ret}
print(rett)
print(json.dumps(rett, ensure_ascii=False))
php code print nothing. How can I solve this problem?
from here, shell_exec is disabled when PHP is running in safe mode.
but you can try this,
<?php
$result;
$resultD;
$arg= $ord." ".$time2_int;
$command = escapeshellcmd('python3 path/to/python_file/decemberlist.py '.$arg);
$result= shell_exec($command);
$resultD=json_decode($result,true);
echo $resultD['song'];
?>
I am working on this project that requires me to upload pictures on PHP, execute the picture on python, fetch the output from python and display it again on PHP.
PHP code:
<?php
$command = shell_exec("python C:/path/to/python/KNNColor.py");
$jadi = json_decode($command);
var_dump($jadi);
?>
Python code:
from PIL import Image
import os
import glob
import cv2
import numpy as np
import matplotlib.pyplot as plt
from skimage import io, color
from scipy.stats import skew
#data train untuk warna
Feat_Mom_M = np.load('FeatM_M.npy')
Feat_Mom_I = np.load('FeatM_I.npy')
Malay_Col_Train = Feat_Mom_M
Indo_Col_Train = Feat_Mom_I
#Data warna
All_Train_Col = np.concatenate((Malay_Col_Train, Indo_Col_Train))
Y_Indo_Col = [0] * len(Indo_Col_Train)
Y_Malay_Col = [1] * len(Malay_Col_Train)
Y_Col_Train = np.concatenate((Y_Malay_Col, Y_Indo_Col))
Train_Col = list(zip(All_Train_Col, Y_Col_Train))
from collections import Counter
from math import sqrt
import warnings
#Fungsi KNN
def k_nearest_neighbors(data, predict, k):
if len(data) >= k:
warnings.warn('K is set to a value less than total voting groups!')
distances = []
for group in data:
for features in data[group]:
euclidean_dist = np.sqrt(np.sum((np.array(features) - np.array(predict))**2 ))
distances.append([euclidean_dist, group])
votes = [i[1] for i in sorted(distances)[:k]]
vote_result = Counter(votes).most_common(1)[0][0]
return vote_result
image_list = []
image_list_pixel = []
image_list_lab = []
L = []
A = []
B = []
for filename in glob.glob('C:/path/to/pic/uploaded/batik.jpg'):
im=Image.open(filename)
image_list.append(im)
im_pix = np.array(im)
image_list_pixel.append(im_pix)
#ubah RGB ke LAB
im_lab = color.rgb2lab(im_pix)
#Pisah channel L,A,B
l_channel, a_channel, b_channel = cv2.split(im_lab)
L.append(l_channel)
A.append(a_channel)
B.append(b_channel)
image_list_lab.append(im_lab)
<The rest is processing these arrays into color moment vector, it's too long, so I'm skipping it to the ending>
Feat_Mom = np.array(Color_Moment)
Train_Set_Col = {0:[], 1:[]}
for i in Train_Col:
Train_Set_Col[i[-1]].append(i[:-1])
new_feat_col = Feat_Mom
hasilcol = k_nearest_neighbors(Train_Set_Col, new_feat_col, 9)
import json
if hasilcol == 0:
#print("Indonesia")
print (json.dumps('Indonesia'));
else:
#print("Malaysia")
print (json.dumps('Malaysia'));
So as you can see, There is only one print command. Shell_exec is supposed to return the string of the print command from python. But what I get on the "var_dump" is NULL, and if I echo $jadi, there's also nothing. Be it using print or the print(json) command
The fun thing is, when I try to display a string from this python file that only consists 1 line of code.
Python dummy file:
print("Hello")
The "Hello" string, shows up just fine on my PHP. So, is shell_exec unable to read many codes? or is there anything else that I'm doing wrong?
I finally found the reason behind this. In my python script there are these commands :
Feat_Mom_M = np.load('FeatM_M.npy')
Feat_Mom_I = np.load('FeatM_I.npy')
They load the numpy arrays that I have stored from the training process in KNN and I need to use them again as the references for my image classifying process in python. I separated them because I was afraid if my PHP page would take too long to load. It'd need to process all the training data, before finally classifying the uploaded image.
But then when I execute my python file from PHP, I guess it returns an error after parsing those 2 load commands. I experimented putting the print command below them, and it stopped showing on PHP. Since it's all like this now, there's no other way than taking the worst option, even if it'd cost me long loading time.
I tested this in the console:
php > var_dump(json_decode("Indonesia"))
php > ;
php shell code:1:
NULL
php > var_dump(json_decode('{"Indonesia"}'))
php > ;
php shell code:1:
NULL
php > var_dump(json_decode('{"Indonesia":1}'))
php > ;
php shell code:1:
class stdClass#1 (1) {
public $Indonesia =>
int(1)
}
php > var_dump(json_decode('["Indonesia"]'))
php shell code:1:
array(1) {
[0] =>
string(9) "Indonesia"
}
you have to have it wrapped in {} or [] and it will be read into an object or an array.
After an error you can run this json_last_error() http://php.net/manual/en/function.json-last-error.php and it will give you an error code the one your's returns should be JSON_ERROR_SYNTAX
I have the code in PHP side:
function apiTest($arga){
if (!isset($arga['name'])) {
printf("Missing required parameter 'name'\n");
return RET_PARAMSMISSING;
}
$retVal = 'testing';
printf("JSON:\n%s",json_wrapper($retVal)
return RET_NOERROR;
}
case 'testing':
$retval = apiTest($arga);
break;
In Browser running this script locahost/testing.php. Here is printing correctly:
TIDBITS API
C:/wamp/www/testing.php running # Sat, 10 Dec 2016 14:08:06
Executing function 'Test'
JSON:
testing
RETURN: 0
Here is json_wrapper printing the output correctly.
This Json_wrapper value how to pass to python function.
I have tried using the below code, but it's not working.
def inventory(self, **kwargs):
return self.tidbitsapi('testing')
I need to get the PHP variable to python side and perform some operation and once pass to return value true/false to PHP.
Try using a system call:
from php:
system('C:/pythonXX/python.exe pathToPythonModule.py "' . arg . '"')
from python:
import sys
if __name__ == '__main__':
if len(sys.argv) > 2:
varFromPhp = sys.argv[2]
print(varFromPhp)
else: print("Didn't get parameter")
This question already has answers here:
Running a Python script from PHP
(10 answers)
Closed 7 years ago.
I have a python script that I would like to run from PHP. This is my PHP script:
$data = array('as', 'df', 'gh');
// Execute the python script with the JSON data
$result = shell_exec('python /path/to/myScript.py ' . escapeshellarg(json_encode($data)));
// Decode the result
$resultData = json_decode($result, true);
// This will contain: array('status' => 'Yes!')
var_dump($resultData);
And this is my Python script:
import sys, json
# Load the data that PHP sent us
try:
data = json.loads(sys.argv[1])
except:
print "ERROR"
sys.exit(1)
# Generate some data to send to PHP
result = {'status': 'Yes!'}
# Send it to stdout (to PHP)
print json.dumps(result)
I would like to be able to exchange data between PHP and Python, but the above error gives the output:
ERROR NULL
Where am I going wrong ?
:::::EDIT::::::
I ran this:
$data = array('as', 'df', 'gh');
// Execute the python script with the JSON data
$temp = json_encode($data);
$result= shell_exec('C:\Python27\python.exe test.py ' . "'" . $temp . "'");
echo $result;
I am getting No JSON object could be decoded
On my machine, the code works perfectly fine and displays:
array(1) {
'status' =>
string(4) "Yes!"
}
On the other hand, you may make a few changes to diagnose the issue on your machine.
Check the default version of Python. You can do this by running python from the terminal. If you see something like:
Python 2.7.6 (default, Mar 22 2014, 22:59:56)
[GCC 4.8.2] on linux2
Type "help", "copyright", "credits" or "license" for more information.
>>>
you're fine. If you see that you are running Python 3, this could be an issue, since your Python script is written for Python 2. So:
Python 3.4.0 (default, Apr 11 2014, 13:05:11)
[...]
should be a clue.
Again from the terminal, run python myScript.py "[\"as\",\"df\",\"gh\"]". What do you see?
{"status": "Yes!"}
is cool. A different response indicates that the issue is probably with your Python script.
Check permissions. How do you run your PHP script? Do you have access to /path/to/? What about /path/to/myScript.php?
Replace your PHP code by:
<?php
echo file_get_contents("/path/to/myScript.php");
?>
Do you get the actual contents?
Now let's add a few debugging helpers in your PHP code. Since I imagine that you are not using a debugger, the simplest way is to print debug statements. This is OK for 10-LOC scripts, but if you need to deal with larger applications, invest your time in learning how to use PHP debuggers and how do use logging.
Here's the result:
/path/to/demo.php
<?php
$data = array('as', 'df', 'gh');
$pythonScript = "/path/to/myScript.py";
$cmd = array("python", $pythonScript, escapeshellarg(json_encode($data)));
$cmdText = implode(' ', $cmd);
echo "Running command: " . $cmdText . "\n";
$result = shell_exec($cmdText);
echo "Got the following result:\n";
echo $result;
$resultData = json_decode($result, true);
echo "The result was transformed into:\n";
var_dump($resultData);
?>
/path/to/myScript.py
import sys, json
try:
data = json.loads(sys.argv[1])
print json.dumps({'status': 'Yes!'})
except Exception as e:
print str(e)
Now run the script:
cd /path/to
php -f demo.php
This is what I get:
Running command: python /path/to/myScript.py '["as","df","gh"]'
Got the following result:
{"status": "Yes!"}
The result was transformed into:
array(1) {
'status' =>
string(4) "Yes!"
}
yours should be different and contain a hint about what is happening.
I got it to work by adding quotes around the argument!
Like so:
<?php
$data = array('as', 'df', 'gh');
$temp = json_encode($data);
echo shell_exec('python myScript.py ' . "'" . $temp . "'");
?>
I have a python script that returns a json object. Say, for example i run the following:
exec('python /var/www/abc/abc.py');
and it returns a json object, how can i assign the json object as a variable in a php script.
Example python script:
#!/usr/bin/python
import sys
def main():
data = {"Fail": 35}
sys.stdout.write(str(data))
main()
Example PHP script:
<?php
exec("python /home/amyth/Projects/test/variable.py", $output, $v);
echo($output);
?>
The above returns an empty Array. Why so ?
I want to call the above script from php using the exec method and want to use the json object returned by the python script. How can i achieve this ?
Update:
The above works if i use another shell command, For Example:
<?php
exec("ls", $output, $v);
echo($output);
?>
Anyone knows the reason ?
If the idea is you'll have a Python script which prints JSON data to standard out, then you're probably looking for popen.
Something like...
<?php
$f = popen('python /var/www/abc/abc.py', 'r');
if ($f === false)
{
echo "popen failed";
exit 1;
}
$json = fgets($f);
fclose($f);
...will grab the output into the $json variable.
As for your example Python script, if the idea is you're trying to convert the Python dictionary {"tests": "35"} to JSON, and print to standard out, you need to change loads to dumps and return to print, so it looks like...
import simplejson
def main():
data = simplejson.dumps({"tests": "35"})
print data
main()