I cannot import from excel to Mysql with php

Question

with this update, It no shows any error, and when I sent the print, it shows the information that I want to add into the database but it does not add it...

if(isset($_POST["submit_file"]))
$file = $_FILES["file"]["tmp_name"];
$file_open = fopen($file,"r");
$count = 0; 
while(($csv = fgetcsv($file_open, 100, ",")) !== false) 
{ 
      $agente = isset($csv[0])? $csv[0]: null;
    $revenue = isset($csv[1])? $csv[1]: null;
    $css = isset($csv[2])? $csv[2]: null;
    $rating = isset($csv[3])? $csv[3]: null;
    $absent = isset($csv[4])? $csv[4]: null;
    $score = isset($csv[5])? $csv[5]: null;
    $count++;
    if($count>1){ 
      $query = "INSERT INTO metricas values ('$agente', $revenue, $css, $rating, $absent, $score)";
  print $query;  }

     }

?>

Possible duplicate of [PHP parse/syntax errors; and how to solve them?](https://stackoverflow.com/questions/18050071/php-parse-syntax-errors-and-how-to-solve-them) — Jens, Jun 20 '18 at 08:09
`$agente[$csv[0]] = isset($csv[1])? $csv[1]: null;` seems wrong to me. For the current query to work, it should be `$agente = isset($csv[1])? $csv[1]: null;`. — Romeo Sierra, Jun 20 '18 at 08:10
@sam that's a really pretty name for curly brackets, I'm stealing that from you — Loek, Jun 20 '18 at 08:11
@RomeoSierra ti show me now Parse error: syntax error, unexpected end of file in /home/wc3al7rafj3e/public_html/moffatutah.net/test1/import_file.php on line 22 — OcYriZ jImEneZ, Jun 20 '18 at 08:13
You have a missing curly brace. One more curly brace should come. I have proposed an edit to your question. Please accept and check it... — Romeo Sierra, Jun 20 '18 at 08:15

score 0 · Answer 1 · edited Jun 20 '18 at 09:08

0

Learn to indent your code properly, that will help you locate problems with quoting, parenthses, brackets, and braces.

You can also use [php Code Checker][https://phpcodechecker.com/] to locate parsing errors in your code.

<?php
require_once 'connect.php';

if(isset($_POST["submit_file"]))
$file = $_FILES["file"]["tmp_name"];
$file_open = fopen($file,"r");
$count = 0; 
while(($csv = fgetcsv($file_open, 100000, ",")) !== false) {
    //$agente[$csv[0]] = isset($csv[1])? $csv[1]: null; // This is wrong in terms of semantics. It should be as per the following line
    $agente = isset($csv[1])? $csv[1]: null;
    // Below lines also should be changed. Hence did change them.
    $revenue = isset($csv[2])? $csv[2]: null;
    $css = isset($csv[3])? $csv[3]: null;
    $rating = isset($csv[4])? $csv[4]: null;
    $absent = isset($csv[5])? $csv[5]: null;
    $score = isset($csv[6])? $csv[6]: null;
    $count++;
    if($count>1){ 
        $query = "INSERT INTO metricas values (null, '$agente', $revenue, $css, $rating, $absent, $score)";
    }
}   //  Added this to close the while loop.

?>

EDIT Modified the code to take out Warning: Array to String conversion.

edited Jun 20 '18 at 09:08

Romeo Sierra

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answered Jun 20 '18 at 08:22

Sloan Thrasher

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I did the test and it shows me Notice: Array to string conversion in /home/wc3al7rafj3e/public_html/moffatutah.net/test1/import_file.php on line 32 – OcYriZ jImEneZ Jun 20 '18 at 08:47
Which line is 32? – Sloan Thrasher Jun 20 '18 at 08:58
$query = "INSERT INTO metricas values (null, '$agente', $revenue, $css, $rating, $absent, $score)"; – OcYriZ jImEneZ Jun 20 '18 at 08:59
Well, duh! In your INSERT statement, you use variables that are arrays. Look at your assignment statements above that line. ```$agente``` is an array. ```$agente[$csv[0]]``` is the value. Why are you assigning to ```$agente[$csv[0]]``` instead of ```$agente```?? – Sloan Thrasher Jun 20 '18 at 09:04
@romeo sierra and Sloan Thrasher I've change as you said – OcYriZ jImEneZ Jun 20 '18 at 09:19
Well yes because I have not any error but, I sent print $query and it shows the info but is not added into the database... – OcYriZ jImEneZ Jun 20 '18 at 09:25

I cannot import from excel to Mysql with php

1 Answers1