MySQL Nested Statement

Question

if ($resultdetails == FALSE) {
    $querynewid = "select * from customer_det order by ID desc limit 1";
    $resultnewid = mysql_query($querynewid);
    while ($row=mysql_fetch_row($resultnewid)) {
        $uid = $row['id'];
        echo $uid;
    }
}

Is this a valid nested statement to make? It won't echo $uid for some reason?

First step: Exchange that 'echo' with a 'var_dump'. Also, 'var_dump' $row. Reason:Your query most likely is failing for some reason, or it simply can't find a thing. The code itself looks valid. — ATaylor
– ATaylor, Commented Nov 2, 2012 at 14:18
If $resultnewid equals 'NULL' (check with var_dump), that means your query is faulty. In that case, echo mysql_error. This will give you a hint as to what's wrong. — ATaylor
– ATaylor, Commented Nov 2, 2012 at 14:20
After ATaylor's comments, you can print query and run in mysql client to see if there is something for you. — Riz
– Riz, Commented Nov 2, 2012 at 14:20
Not logic part: limit 1 and while ($row=mysql_fetch_row($resultnewid)). There is only 1 result.. — Ron van der Heijden
– Ron van der Heijden, Commented Nov 2, 2012 at 14:21
I am guessing $resultdetails is never false, can you provide that part? — Ryan B
– Ryan B, Commented Nov 2, 2012 at 14:21

Community · Accepted Answer · 2023-11-17 20:08:53Z

1

You should use mysql_fetch_assoc instead of mysql_fetch_row. The last one returns a numerical indexed-array, so id is not a key in that result set. mysql_fetch_assoc returns an associative array, so the field name is also the name of the key. By the look of your code, that is what you want.

See the difference here and here.

edited Nov 17, 2023 at 20:08

CommunityBot

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answered Nov 2, 2012 at 14:21

Oldskool

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2 Comments

smada Over a year ago

Nice @Oldskool, this pulled the ID perfectly.

ATaylor Over a year ago

My bad, I totally failed to see that he used fetch_row instead of fetch_assoc.

Alex · Accepted Answer · 2012-11-02 15:55:28Z

1

How I usually handle this in php is below:

if ($resultdetails == FALSE) {
    $querynewid = mysqli_query($dbc,"select * from customer_det order by ID desc limit 1";) or die (mysql_error());

    while ($row = mysqli_fetch_array($querynewid)) {
        $uid = $row['id'];
        echo $uid;
    }
}

A couple things to know is the $dbc is my database connection and that the mysql_error will let you know if its the query that itself that is broken

edited Nov 2, 2012 at 15:55

answered Nov 2, 2012 at 14:27

Alex

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2 Comments

Alex Over a year ago

Sorry, the $row = mysqli_fetch_array($resultnewid) should be $row = mysqli_fetch_array($querynewid )

Oldskool Over a year ago

Edited the answer to include that instead.

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