E-COM-NET

Question

I am currently using a JSON encoded array to display the users in my database for an auto-suggest feature.

It looks something like this:

$sth = mysql_query("SELECT id, name FROM users");

$json = array();

    while($row = mysql_fetch_assoc($sth)) {
        $json['name'] = $row['name'];
        $json['id'] = $row['id'];
        $data[] = $json;
    }

print json_encode($data);

This returns:

[{"id":"81","name":"John Doe"},{"id":"82","name":"Jane Doe"}]

My question is somewhat 2-fold:

First, how would I manually add an additional object to this output? For example, let's say I wanted to add: {"id":"444","name":"A New Name"}

Thus, it'd look like:

[{"id":"81","name":"John Doe"},{"id":"82","name":"Jane Doe"},{"id":"444","name":"A New Name"}]

Second, let's say I also wanted to add more objects to the array from a separate table as well, such as:

$sth = mysql_query("SELECT id, title FROM another_table");

$json = array();

    while($row = mysql_fetch_assoc($sth)) {
        $json['name'] = $row['title'];
        $json['id'] = $row['id'];
        $data[] = $json;
    }

print json_encode($data);

This way I could have both tables populated in the JSON array, thus, showing up as additional options in my autosuggest.

Hopefully this makes sense, as I've tried hard to articulate what I am trying to accomplish.

Thanks!

And why don't you add them to your array before you do the json_encode? That would make more sense to me. — Aug 17 '10 at 15:46

meder 83.7k 25 213 305 · Accepted Answer · Aug 17 '10 at 15:46

Just keep pushing to the $data array.

$json = array();

    while($row = mysql_fetch_assoc($sth)) {
        $json['name'] = $row['name'];
        $json['id'] = $row['id'];
        $data[] = $json;
    }

$custom = array('name'=>'foo', 'id' => 'bar');
$data[] = $custom;

Then at the very end, do your json_encode. Assuming you're not referring to merging it in the JS itself with multiple ajax calls.

And if you have separate scripts, combine them in one php page.

Thanks Meder. Easier than I thought it'd be. – Dodinas Aug 17 '10 at 17:04 — Aug 17 '10 at 17:04

Colin Fine 2,390 8 17 · Answer 2 · Aug 17 '10 at 15:46

up vote 0 down vote

You could edit the JSON (text), but it's much easier to modify the array before you encode it. Or am I missing something?

answered Aug 17 '10 at 15:46

Colin Fine
2,390 8 17

add a comment

score 0 · Answer 3

I'm not an expert in any of these fields, but I'll try and see if I can help. Try one of these:

Option 1 (I don't know how unions work in mysql):

$sth = mysql_query("SELECT id, name FROM users union SELECT id, name FROM (SELECT id, title as name from another_table) as T2"); 


    $json = array(); 

        while($row = mysql_fetch_assoc($sth)) { 
            $json['name'] = $row['name']; 
            $json['id'] = $row['id']; 
        }

    $json['name'] = 'A new name';
    $json['id'] = '444';
    $data[] = $json; 

    print json_encode($data);

I've never done PHP, so I'm making assumptions. I've also never used MySql, so there's more assumptions.

Option 2:

$sth = mysql_query("SELECT id, name FROM users"); 


    $json = array(); 

        while($row = mysql_fetch_assoc($sth)) { 
            $json['name'] = $row['name']; 
            $json['id'] = $row['id']; 
        }

$sth = mysql_query("SELECT id, title from another_table"); 


        while($row = mysql_fetch_assoc($sth)) { 
            $json['name'] = $row['title']; 
            $json['id'] = $row['id']; 
        }

    $json['name'] = 'A new name';
    $json['id'] = '444';
    $data[] = $json; 

    print json_encode($data);

Hope this helps.

meder 83.7k 25 213 305 · Accepted Answer · Aug 17 '10 at 15:46

up vote 8 down vote accepted

Just keep pushing to the $data array.

$json = array();

    while($row = mysql_fetch_assoc($sth)) {
        $json['name'] = $row['name'];
        $json['id'] = $row['id'];
        $data[] = $json;
    }

$custom = array('name'=>'foo', 'id' => 'bar');
$data[] = $custom;

Then at the very end, do your json_encode. Assuming you're not referring to merging it in the JS itself with multiple ajax calls.

And if you have separate scripts, combine them in one php page.

answered Aug 17 '10 at 15:46

meder
83.7k 25 213 305

Thanks Meder. Easier than I thought it'd be. – Dodinas Aug 17 '10 at 17:04

add a comment

Colin Fine 2,390 8 17 · Answer 5 · Aug 17 '10 at 15:46

up vote 0 down vote

You could edit the JSON (text), but it's much easier to modify the array before you encode it. Or am I missing something?

answered Aug 17 '10 at 15:46

Colin Fine
2,390 8 17

add a comment

score 0 · Answer 6

I'm not an expert in any of these fields, but I'll try and see if I can help. Try one of these:

Option 1 (I don't know how unions work in mysql):

$sth = mysql_query("SELECT id, name FROM users union SELECT id, name FROM (SELECT id, title as name from another_table) as T2"); 


    $json = array(); 

        while($row = mysql_fetch_assoc($sth)) { 
            $json['name'] = $row['name']; 
            $json['id'] = $row['id']; 
        }

    $json['name'] = 'A new name';
    $json['id'] = '444';
    $data[] = $json; 

    print json_encode($data);

I've never done PHP, so I'm making assumptions. I've also never used MySql, so there's more assumptions.

Option 2:

$sth = mysql_query("SELECT id, name FROM users"); 


    $json = array(); 

        while($row = mysql_fetch_assoc($sth)) { 
            $json['name'] = $row['name']; 
            $json['id'] = $row['id']; 
        }

$sth = mysql_query("SELECT id, title from another_table"); 


        while($row = mysql_fetch_assoc($sth)) { 
            $json['name'] = $row['title']; 
            $json['id'] = $row['id']; 
        }

    $json['name'] = 'A new name';
    $json['id'] = '444';
    $data[] = $json; 

    print json_encode($data);

Hope this helps.

Add Additional Objects to JSON Encoded Array

3 Answers

3 Answers

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