Solution for PHP Notice: Undefined index: internalid [closed]

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Solution for PHP Notice: Undefined index: internalid [closed]

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#1 by (1 votes)

2

I do not understand what is wrong with this code

//$expense_categories = explode(chr(13), $row[expense_categories]);
$categoryResult=mysqli_query($db,"SELECT category_id,category FROM expense_categories order by category_id");
if(mysqli_num_rows($categoryResult) > 0){
//echo mysqli_num_rows($categoryResult)."<br/><br/>\n";
    while($catrow = mysqli_fetch_assoc($categoryResult)){
        //echo $catrow["internalid"],"  ",$catrow["category"]."<br/>";
        $expense_categories[$catrow["internalid"]]=$catrow["category_id"]; //Linea 14
    }
}

Error in log:

PHP Notice: Undefined index: internalid in /var/www/misitio.org/cc/config.inc on line 14, referer: link

php array indices

asked by Martini002 30.03.2017 в 20:25

source

1 answer

Create a web service with java that returns a query from bd in json [closed] Run at the beginning a script developed with nodejs in ubuntu server 16?

score 1 · Accepted Answer

The Error is simple, you can not access the Index internalid given that it is not part of your query SELECT , you can only access the Indices that you call in your query. That is, $catrow["category "] and $catrow["category_id"]