MySQLi not showing any error

Question

I have been trying to solve this for hours now and just can't figure out what I am doing wrong here. The loop won't cycle (nothing is being echo'd), but when I run the query in phpmyadmin it works fine. Also the database info is fine. Unfortunately mysqli doesn't output any error at all...

Any help is greatly appreciated!

<?php
ini_set('display_errors', '1');
error_reporting(E_ALL);
set_time_limit(0);

$db = new mysqli('localhost', 'xxx', 'xxx', 'xxx');

if($db->connect_errno > 0){
    die('Unable to connect to database [' . $db->connect_error . ']');
}

$sql = <<<SQL
    SELECT * FROM 
    `zips`
SQL;


if(!$result = $db->query($sql,MYSQLI_STORE_RESULT)){
    echo  $result->num_rows; // nothing outputted here
    while($row = $result->fetch_assoc()){
        $zip    = $row['zip'];        
        $url = "http://www.webservicex.net/uszip.asmx/GetInfoByZIP?USZip=".$zip;

        $ch = curl_init();
        curl_setopt($ch, CURLOPT_URL, $url);
        curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);

        $result = curl_exec($ch);

        curl_close($ch);

        $xmlobj = simplexml_load_string($result);
        $city = $xmlobj->Table->CITY;
        $state = $xmlobj->Table->STATE;
        $timezone = $xmlobj->Table->TIME_ZONE;
        $areacode = $xmlobj->Table->AREA_CODE;

        echo "city:" . $city;die();

        $sql2 = <<<SQL
            INSERT INTO zips
            (zip,city,state,timezone,areacode)
            VALUES('{$zip}','{$city}','{$state}', '{$timezone}', '{$areacode}')
SQL;
        if(!$result2 = $db->query($sql2)){
            die('There was an error running the query [' . $db->error . ']');
        }
    }
} else {
        die('There was an error running the query [' . $db->error . ']');
}

this should work. How are you using this as, `http://localhost` or `file:///`? — Funk Forty Niner, Oct 13 '16 at 14:34
looks like your username and password are all x's, other then that I would turn on display errors `ini_set('display_errors', 1);` — ArtisticPhoenix, Oct 13 '16 at 14:34
@ArtisticPhoenix x's are used to hide the real credentials and PHP is a loosely typed language. — Your Common Sense, Oct 13 '16 at 14:36
you were close Fred `f(!$result2 = $db->query($sql2))` assignment, should be `if(false === ( $result2 = $db->query($sql2)))` — ArtisticPhoenix, Oct 13 '16 at 14:37
@ArtisticPhoenix actually, I deleted my comment about that. But their way to use that file is unknown. — Funk Forty Niner, Oct 13 '16 at 14:37
You are vulnerable to [sql injection attacks](http://bobby-tables.com). — Marc B, Oct 13 '16 at 14:38
@Machavity: what's wrong with heredocs? Changing how a string is build doesn't change the string. `$x = 'foo';` and `$x = <<< ... foo` both produce the exact same string. — Marc B, Oct 13 '16 at 14:38
its a double negative, if not false..... but its pretty early for me so I may be way off. lol — ArtisticPhoenix, Oct 13 '16 at 14:38
@ArtisticPhoenix or `if($result` no idea why they're using a `!` here. — Funk Forty Niner, Oct 13 '16 at 14:40
@Fred-ii- I was thinking how its common to do `while( false !== ( $row = $qeury->fetch() ))` assignment and comparison. — ArtisticPhoenix, Oct 13 '16 at 14:42
@Machavity it's entirely a matter of taste, and mentioning such matters just polluting comments. — Your Common Sense, Oct 13 '16 at 14:43
The only issue whit heredoc is that the ending tag needs to be on the line all by itself, hard to tell on here if there is any spaces after the `SQL;` such as `SQL;\s` would break it. Personally I like Heredoc for multi line strings. But I only use the NewDoc for my sql. `<<<'SQL'` But I don't be messing with the sql injection either ... lol — ArtisticPhoenix, Oct 13 '16 at 14:45
This will screw the pooch as they say `echo "city:" . $city;die();` Die die die. — ArtisticPhoenix, Oct 13 '16 at 14:47
Ok, the error is just a misplaced ! operator. I closed thie cuestion with a link to a better technique that is not prone to errors of this sort. — Your Common Sense, Oct 13 '16 at 14:48
@Johan for this very code just remove first ! mark from it. But better yet, check the linked answer. — Your Common Sense, Oct 13 '16 at 14:48
@YourCommonSense I can't see why you closed the question, seeing the OP states: *"Unfortunately mysqli doesn't output any error at all"* - They are using error reporting and error handling on the db/queries. Personally, the question should be reopened. — Funk Forty Niner, Oct 13 '16 at 14:49
@ArtisticPhoenix this is intentional die added for debugging. — Your Common Sense, Oct 13 '16 at 14:49
@YourCommonSense - this die `echo "city:" . $city;die();` is not for debugging. I get what your saying but the sql is not even being ran because of it. — ArtisticPhoenix, Oct 13 '16 at 14:50
One last question, (first time I ask a question here) how can I close this question? — Johan, Oct 13 '16 at 14:52
@YourCommonSense Pretty ironic in your answer for the duplicate you posted and I quote you: *"use die() or echo or any other function to print the error message on screen unconditionally. PHP can echo it all right already, no assistance required."* - the OP did use `die()`, just shouldn't have used it in the first place. I reopened the question. — Funk Forty Niner, Oct 13 '16 at 14:58

ArtisticPhoenix · Answer 1 · 2016-10-13T15:12:17.280

2

Here is your issue Mate

  echo "city:" . $city;die();  ///script stops.

     .....

 if(!$result2 = $db->query($sql2)){

DIE - which will prematurely end your script before running the query inside this block.

Also Technically you can get away with this

    if(!$result2 = $db->query($sql2)){

But it's very hard to read ( a double negative ) I would change it to this way

 if(false === ($result2 = $db->query($sql2))){

As you have it you are saying if(!false == true ) which will pass on the false value and output the error ect.

AS Your Common Sense mentioned you may not be in this block, but without knowing what the result of the first query is, ( it would need to be false to be running in the posted block of code ) it's hard to say. It could be that that first query should be false?

 if(!$result = $db->query($sql,MYSQLI_STORE_RESULT)){
       .....

To explain a bit more, the above line would need to return false, then with the ! not it would be true, for the code in the if block below it to even run. However in that case ( if it was not false ) this output would be generated

   ....
} else {
    die('There was an error running the query [' . $db->error . ']');
}

More likely its a combination of errors.

edited Oct 13 '16 at 15:12

answered Oct 13 '16 at 14:48

ArtisticPhoenix

21,464
2
24
38

Not my downvote here, but I can make a real wild guess as to where that comes from. – Funk Forty Niner Oct 13 '16 at 14:51
The OP clearly stated that they don't see any output. While this code shoud let out at least the word "city:" – Your Common Sense Oct 13 '16 at 14:51
@YourCommonSense so why didn't you respond to [my comment](http://stackoverflow.com/questions/40023929/mysqli-not-showing-any-error#comment67325727_40023929) about your closing the question? which should be reopened. IMHO. If you don't then I will and don't go on meta about it when I do, you will get shot down for it or flag the question. – Funk Forty Niner Oct 13 '16 at 14:52
That's because the die preceeds the sql query, in other words the script is ending without executing the sql because of the misplaced die. Ergo no sql is being executed and therefor no error or success is possible. – ArtisticPhoenix Oct 13 '16 at 14:52
@ArtisticPhoenix Seems like I've fallen onto deaf ears with that guy. But pretty sure he will comment once I reopen the question if he doesn't. – Funk Forty Niner Oct 13 '16 at 14:55
@ArtisticPhoenix did you notice *another* SQL query, near the top of the script, by chance? – Your Common Sense Oct 13 '16 at 14:57
Nope didn't notice that at all. Just being honest, but do you honestly think that die belongs there? – ArtisticPhoenix Oct 13 '16 at 14:57
then you may reconsider your conclusions now. taking into account echo `$result->num_rows;` line as well – Your Common Sense Oct 13 '16 at 14:58
1

Not likely, I feel I am still correct. None of the other code would even execute if that was the case. Sense that is the only code there one must assume that some out put is generated by that block, given the echo city its a safe bet. Either they get the `city` as output or this error `die('There was an error running the query [' . $db->error . ']');` From the first queries else block. as they said no errors, I am safe in my assumption. – ArtisticPhoenix Oct 13 '16 at 15:00
So you *insists* that die will be executed *prior* `$result->num_rows;`? Nuff said. – Your Common Sense Oct 13 '16 at 15:01
The OP said no errors are output, the else for the first query outputs an error, so if the first query failed they would have an error output. Therefore its a very safe bet that the issue lies within the code block that was posted. My conclusion is irrefutable. I'm not always right but I'm never wrong.... – ArtisticPhoenix Oct 13 '16 at 15:02
@YourCommonSense - don't get me wrong I see what you are saying but there is no way to know if that first query should return no results. It's possible it shouldn't. Anyway I updated my answer to reflect that. – ArtisticPhoenix Oct 13 '16 at 15:13

MySQLi not showing any error

1 Answers1