While loop together with foreach - php

Question

I'm really stuck trying to resolve what should be quite simple.
I Have this
<?php
$json = json_decode('{
"33540116":
{"person":
{"name":"John", "age":"36"}},
"33541502":
{"person":
{"name":"Jack", "age":"23"}}
}
');
$id = array('33540116', '33541502');
foreach($id as $id) {
echo $json->$id->person->{'name'}. '<br />';
echo $json->$id->person->{'age'}. '<br />';
}
?>
So the code is decoding a json string then using foreach to echo each result.
This json file is rather large and I'm only interested in certain records that match the id's stored in a mysql table.
To do this I have replaced the id array string above with mysql select statement.
<?php
$json = json_decode('{
"33540116":
{"person":
{"name":"John", "age":"36"}},
"33541502":
{"person":
{"name":"Jack", "age":"23"}}
}
');
$result = mysql_query("SELECT id FROM people");
$row = mysql_fetch_array($result);
$id = array($row['id']);
foreach($id as $id) {
echo $json->$id->person->{'name'}. '<br />';
echo $json->$id->person->{'age'}. '<br />';
}
?>
Although this works, it only gives me 1 result.
What I really need is to loop through the results.
Unfortunately I don't know how to construct a while loop together with foreach.
I will greatly appreciate your assistance.
UPDATE (extra question)
Thanks everyone. You have helped me solve the problem.
However, I have another question that relates to this matter.
I mentioned above that I merely wanted to echo the results.
But this isn't exactly true.
What I really want to do is update the same mysql table with the results retreived from the json file.
I have a table called people with fields id, name and age.
How can I update this table with these results?
Thanks again.

Answer 1

mysql_fetch_array only fetches one row at a time. You can use a while loop to continue fetching rows. The mysql_fetch_array function returns false once the whole result set has been fetched, so that will cause the while loop to terminate as desired.
Also, I removed the foreach loop on $ids. Since there will only be one element in the array it's unnecessary to put the code in a loop.
<?php
$json = json_decode('{
"33540116":
{"person":
{"name":"John", "age":"36"}},
"33541502":
{"person":
{"name":"Jack", "age":"23"}}
}
');
$result = mysql_query("SELECT id FROM people");
while ( ( $row = mysql_fetch_array($result) ) ) {
$id = $row['id'];
echo $json->$id->person->{'name'}. '<br />';
echo $json->$id->person->{'age'}. '<br />';
}
?>

Answer 2

Try this:
while($row = mysql_fetch_array($result)) {
$id = array($row['id']);
foreach($id as $id) {
echo $json->$id->person->{'name'}. '<br />';
echo $json->$id->person->{'age'}. '<br />';
}
}

Answer 3

You are using this :
$row = mysql_fetch_array($result);
This will only fetch one row from the database.
If you want to fetch more than one row, you have to call mysql_fetch_array() in a loop :
while ($row = mysql_fetch_array($result)) {
// Work with the current row
}
In your case, you'd have something like this :
while ($row = mysql_fetch_array($result)) {
$id = $row['id'];
echo $json->$id->person->{'name'}. '<br />';
echo $json->$id->person->{'age'}. '<br />';
}

Answer 4

You do not need foreach loop the while is just enough
$result = mysql_query("SELECT id FROM people");
$id = array($row['id']);
while($row = mysql_fetch_array($result))
{
echo $json->$row['id']->person->{'name'}. '<br />';
echo $json->$row['id']->person->{'age'}. '<br />';
}

Answer 5

You have a couple of issues. Firstly, you shouldn't be using foreach($id as $id) since you are using the same variable for both. Instead it should be foreach($ids as $id). Secondly, you can get a list of the ids and echo out the correct json values as follows
<?php
$json = json_decode('{
"33540116":
{"person":
{"name":"John", "age":"36"}},
"33541502":
{"person":
{"name":"Jack", "age":"23"}}
}
');
$result = mysql_query("SELECT id FROM people");
$ids = array();
while($row = mysql_fetch_array($result)) {
$ids[] = $row['id'];
}
foreach($ids as $id) {
echo $json->$id->person->{'name'}. '<br />';
echo $json->$id->person->{'age'}. '<br />';
}
?>
Hopefully that will solve it