Array to String Error in Mysql using PHP

Question

I have an Android application that sends a Json as a string to the server in PHP. I've tested with var_dump to verify that the data was being passed correctly and everything is ok. The problem is that when I try to access json and assign values from a nested array of the main object to an array in PHP, I got an error when I try to include this array in Mysql. I've tested only PHP and MySQL and everything is working perfectly.

if  (!empty($_POST)){
   $info = file_get_contents('php://input');
   $json = json_decode($info, true);
   $login= "";
   $useExercise = array();
  //var_dump($info);

   foreach($json['Patient'][0] as $name){
      $login = $name;
   }


   foreach ($json['Patient'][1]as $exercise){
       $useExercise[] = array($exercise);
   }

for ($i=0; sizeOf($useExercise) > $i; $i++){

    $exercicies= mysqli_fetch_array($order);
    $sql1 = ("UPDATE patient_exercise
             INNER JOIN patients ON (patient_exercise.idpatient =  patients.ID)
             INNER JOIN exercises ON (exercises.idexercises  = patient_exercise.idexercise)
             SET patient_exercise.use_exercise=$useExercise[$i]     
             WHERE patient_exercise.idexercise= {$exercises['idexercise']} AND
                   patients.ID=(SELECT c.ID FROM (SELECT * FROM patients ) as c 
                                 WHERE c.login_patients  = '$login');");

    mysqli_query($connect, $sql1);


   }
}

the error occurs on SET patient_exercise.use_exercise=$useExercise[$i]

And my Json is:

{"Paciente":[{"Nome":"Rafael"},
           {"Exercicios":[{"0":"1"},
                        {"1":"0"},
                        {"2":"0"}]}]
}

What could be wrong?

Rodrigo Ferraz · Accepted Answer · 2018-02-02 23:57:13Z

1

So, I've made a mistake by not capturing the data from $json properly. I've also changed my json object. What I did was this:

 $result= array();
    foreach($json as $patients){
        foreach ($patients as $key=>$value){
            foreach ($value as $a=>$b)          
                    $result[] = $b;     
        }       
    }

    for ($i=0; sizeOf($result)>$i;$i++){
        if($i ==0){
            $login = $result[$i];
        } 
        else{
            $useExercise[]=$result[$i];
        }
    }

And my modified Json:

{"patient":[{"name":"rafael"},
             {"0":"1","1":"0","2":"0"}
            ]
}

Thank you, Kingsley Mitchell for taking your time to help me! :)

answered Feb 2, 2018 at 23:57

Rodrigo Ferraz

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1 Comment

Rodrigo Ferraz Over a year ago

Sorry. I just did it! :D

Kingsley Mitchell · Accepted Answer · 2018-02-02 04:57:31Z

0

Think your main problem is you are making it to complex by jumping steps instead of doing it step by step.

Make it simple because it looks confusing and it is hard to correct because your context and naming conventions are not easy to understand.

Try this format : $SQL = "UPDATE isignup SET password='". $NPass ."' WHERE custID='$num'";

edited Feb 2, 2018 at 4:57

answered Feb 2, 2018 at 4:32

Kingsley Mitchell

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26 Comments

Rodrigo Ferraz Over a year ago

Sorry, I've edited my question. What steps could I add so you can understand it better?

Kingsley Mitchell Over a year ago

You are updating multiple rows in a database at once, try splitting it up into one-liners .

Kingsley Mitchell Over a year ago

Joins are notorious of problems

Kingsley Mitchell Over a year ago

Take out the joins and make it without the joins first.

Rodrigo Ferraz Over a year ago

But Im using a FOR. How am I adding multiple rows at once? I'm using each element from an array and adding each one of them at a time. At least from what I understand from the FOR clause.

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