nodirtyrockstar Posted January 8, 2013 Share Posted January 8, 2013 I'm using PHP 5.3.13. I am just using a very simple query to grab one column of results from a table. I would like to fetch them and then iteratively add them to a dropdown menu. I am trying to understand these methods/functions and seem to be missing something. My research on Google didn't give me any indication of what I'm doing wrong. I start with one of the simplest queries possible, which I know will return a data set as I have tested it in mysqladmin. The problem arises when I try to call the method on the mysqli_result object. $query = "SELECT `artist` FROM `bands`;"; $result = $mysqli->query($query); $bandArr = $result->fetch_all(); The error I'm getting from the above code is: Fatal error: "Call to undefined method mysqli_result::fetch_all()..." I researched this error and read somewhere that you need mysqlnd. Is that true? Do I need to look into my PHP configuration to get this to work? Is it worth it for this task? Then I tried fetch_array... $query = "SELECT `artist` FROM `bands`;"; $result = $mysqli->query($query); $bandArr = $result->fetch_array(MYSQLI_NUM); printf("%s\n%s", $bandArr[0], $bandArr[1]); And the above code for some reason returns an array with only one item, and this error: "Notice: Undefined offset: 1 in..." What am I missing here? Again, all I want is a small result set from one column which can be iterated and each value added to a drop down menu. Thoughts? Suggestions? Thanks in advance for any help you can offer... Quote Link to comment https://forums.phpfreaks.com/topic/272871-mysqli-fetch_all-and-fetch_array-errors/ Share on other sites More sharing options...
scootstah Posted January 8, 2013 Share Posted January 8, 2013 (edited) Try this: $bandArr = $result->fetch_array(MYSQLI_NUM); echo '<pre>' . print_r($bandArr,true) . '</pre>'; What do you get? EDIT: Actually, the undefined offset error is because your query is only returning a single column. Edited January 8, 2013 by scootstah Quote Link to comment https://forums.phpfreaks.com/topic/272871-mysqli-fetch_all-and-fetch_array-errors/#findComment-1404317 Share on other sites More sharing options...
DavidAM Posted January 9, 2013 Share Posted January 9, 2013 fetch_array() will return only one row. You need to use it in a loop to process all of the rows: $query = "SELECT `artist` FROM `bands`;"; $result = $mysqli->query($query); while ($bandArr = $result->fetch_array(MYSQLI_NUM)) { printf("%s\n", $bandArr[0]); } Of course, since you are only selecting one column, there will only be one element in the $bandArr array (it will be element zero). Quote Link to comment https://forums.phpfreaks.com/topic/272871-mysqli-fetch_all-and-fetch_array-errors/#findComment-1404366 Share on other sites More sharing options...
nodirtyrockstar Posted January 10, 2013 Author Share Posted January 10, 2013 I am sure you're right. I'll see if I can get this working. Can anyone answer this for me: what is the issue with fetch_all()? Quote Link to comment https://forums.phpfreaks.com/topic/272871-mysqli-fetch_all-and-fetch_array-errors/#findComment-1404792 Share on other sites More sharing options...
Solution DavidAM Posted January 11, 2013 Solution Share Posted January 11, 2013 The error I'm getting from the above code is: Fatal error: "Call to undefined method mysqli_result::fetch_all()..." I researched this error and read somewhere that you need mysqlnd. Is that true? Do I need to look into my PHP configuration to get this to work? Is it worth it for this task? Can anyone answer this for me: what is the issue with fetch_all()? mysqli_fetch_all() is only available with the native driver. I've never used it (fetch_all(), that is). I've never really wanted to use it. Even for populating dropdown selections, I just use a loop. I don't think you will see enough benefit to justify the time and effort to reconfigure PHP. Quote Link to comment https://forums.phpfreaks.com/topic/272871-mysqli-fetch_all-and-fetch_array-errors/#findComment-1404846 Share on other sites More sharing options...
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