illegal string offset when trying to get array value

Question

I'm just trying to get the max value of a row in my table, I then want to insert that value into another table, so I'm trying to get that max value, and every time I'm trying to do something with it, I get this error "illegal string offset" .

$qry2= "SELECT MAX(buss_id) FROM businesses";
$result= mysqli_query($con,$qry2);
$maxid = mysqli_fetch_assoc ($result);

print_r ($maxid);
foreach($maxid as $individual_data)
{
   //Assign the values
   $maxx = $individual_data['buss_id'];
}

My print_r function prints everything properly, but I just can't access the value I need to manipulate. I'm a beginner, please be kind. Thank you. Array ( [MAX(buss_id)] => 47 )

The array key will be the column name. In this case the column name is MAX(buss_id) so either you need to change your query to SELECT MAX(buss_id) AS buss_id or change your code to get the key as $maxx = $individual_data['MAX(buss_id)']; — GordonM
– GordonM, Commented Feb 24, 2017 at 7:51
@GordonM Illegal string offset 'MAX(max_buss_id)' mith's answer worked, though. Thank you man. — Mar1ak
– Mar1ak, Commented Feb 24, 2017 at 9:21

mith · Accepted Answer · 2017-02-24 07:38:24Z

2

Try using alias

$qry2= "SELECT MAX(buss_id) AS maxid FROM businesses";
$result= mysqli_query($con,$qry2);
$maxid = mysqli_fetch_assoc ($result);

echo $maxid['maxid'];

answered Feb 24, 2017 at 7:38

mith

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Comments

ScaisEdge · Accepted Answer · 2017-02-24 07:38:03Z

1

Youn should use a proper alias for assign the name that you use for accessing eg:

 $qry2= "SELECT MAX(buss_id) as buss_id FROM businesses";

answered Feb 24, 2017 at 7:38

ScaisEdge

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ScaisEdge Over a year ago

@Mar1ak . well if my answer is useful (like others ) you could rate properly ... over 15 you can use upper arrow left to answer for sign as useful

Jomoos · Accepted Answer · 2017-02-24 07:38:16Z

1

Since your query is SELECT MAX(buss_id) FROM businesses, the resulting column name will be MAX(buss_id) as is evident the the print_r result you have shared.

Array ( [MAX(buss_id)] => 47 )

You will have to alias the resulting field name as you want to get it:

$qry2   = "SELECT MAX(buss_id) AS max_buss_id FROM businesses";
$result = mysqli_query($con,$qry2);
$maxid  = mysqli_fetch_assoc ($result);

foreach ($maxid as $individual_data)
{
    //Assign the values
    $maxx = $individual_data['max_buss_id'];
}

answered Feb 24, 2017 at 7:38

Jomoos

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5 Comments

Mar1ak Over a year ago

This still gave me the same error. mith's answer worked, though. Your suggestion would work if $individual_data['max_buss_id']; became $maxid['max_buss_id']; I really don't understand it, since $individual_data should be the same as $maxid in the foreach loop. But having $individual_data['max_buss_id'] doesn't work and $maxid['max_buss_id'] works. Thanks for answering!

Jomoos Over a year ago

@Mar1ak Did u noticed the AS max_buss_id in the SQL query? You might have missed it I think.

Mar1ak Over a year ago

@Jmoos Yes bro I did. Also I just edited my first comment, I meant to say mith's answer. You all suggested the same thing, which is correct but for some reason using the foreach's $key variable AKA $individual_data in my case doesn't work. It's no big deal however. I don't need to foreach anything, I was just trying to assign the value to some variable using foreach. It was already achieved in Mith's answer without the use of foreach. I pointed it out cause it's just weird.

Jomoos Over a year ago

@Mar1ak Hmm. What else is causing the problem then? Anyway, good to know that smith's answer worked and you got it solved :)

Mar1ak Over a year ago

@Jmoos I wish I knew. As far as I understand it. The $individual_data should be the same as the $maxid in the foreach loop. Apparently it's not quite like that. Otherwise it should've worked just like you suggested. Luckily I won't need a foreach loop for my code. Thanks a lot man =)

lazyCoder · Accepted Answer · 2017-02-24 07:47:13Z

-1

@Mar1ak you did only one mistake, change $individual_data['buss_id'] to $individual_data['max'] because in your query you use max() without any alias so max() will return value with column name max

$qry2= "SELECT MAX(buss_id) FROM businesses";
    $result= mysqli_query($con,$qry2);
    $maxid = mysqli_fetch_assoc ($result);

print_r ($maxid);
foreach($maxid as $individual_data)
{
   //Assign the values
   $maxx = $individual_data['max'];}

answered Feb 24, 2017 at 7:47

lazyCoder

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2 Comments

GordonM Over a year ago

That still won't work. The array key name is wrong.

lazyCoder Over a year ago

what you are getting in your array print_r ($maxid); ? update your question with this

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