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PHP, returns a JSON encoded array

$this->load->model('car_model', 'cars');
$result = $this->cars->searchBrand($this->input->post('query'));
$this->output->set_status_header(200);
$this->output->set_header('Content-type: application/json');
$output = array();
foreach($result as $r)
    $output['options'][$r->brandID] = $r->brandName;
print json_encode($output);

Outputs: {"options":{"9":"Audi","10":"Austin","11":"Austin Healey"}}

JS updated:

$(".searchcarBrands").typeahead({
    source: function(query, typeahead) {
        $.ajax({
            url: site_url + '/cars/search_brand/'+query,
            success: function(data) {
                typeahead.process(data);
            },
            dataType: "json"
        });
    },
    onselect: function(item) {
        $("#someID").val(item.id);
    }
});

UPDATE: Uncaught TypeError: Object function (){return a.apply(c,e.concat(k.call(arguments)))} has no method 'process'

If I type just 'A' then typeahead shows me only the first letter of each result (a bunch of A letters). If I type a second letter I see nothing anymore.

I've tried JSON.parse on the data or using data.options but no luck.

What am I doing wrong?

PHP, returns a JSON encoded array

$this->load->model('car_model', 'cars');
$result = $this->cars->searchBrand($this->input->post('query'));
$this->output->set_status_header(200);
$this->output->set_header('Content-type: application/json');
$output = array();
foreach($result as $r)
    $output['options'][$r->brandID] = $r->brandName;
print json_encode($output);

Outputs: {"options":{"9":"Audi","10":"Austin","11":"Austin Healey"}}

JS updated:

$(".searchcarBrands").typeahead({
    source: function(query, typeahead) {
        $.ajax({
            url: site_url + '/cars/search_brand/'+query,
            success: function(data) {
                typeahead.process(data);
            },
            dataType: "json"
        });
    },
    onselect: function(item) {
        $("#someID").val(item.id);
    }
});

UPDATE: Uncaught TypeError: Object function (){return a.apply(c,e.concat(k.call(arguments)))} has no method 'process'

If I type just 'A' then typeahead shows me only the first letter of each result (a bunch of A letters). If I type a second letter I see nothing anymore.

I've tried JSON.parse on the data or using data.options but no luck.

What am I doing wrong?

• javascript
• twitter-bootstrap

Share Improve this question Follow edited Nov 6, 2012 at 15:32 stef asked Nov 6, 2012 at 13:48 stefstef 27.8k3131 gold badges107107 silver badges143143 bronze badges 2

• Are you essentially attempting to convert typeahead to allow a remote data source? – Darrrrrren Commented Nov 6, 2012 at 13:55
• OK, see my answer - I believe my code should help you. – Darrrrrren Commented Nov 6, 2012 at 14:00

Add a ment  | 

3 Answers 3

Sorted by: Reset to default 3

I've been battling this for the last day with Bootstrap 2.2.1. No matter what I did, it would not work. For me, I always got the process undefined error unless I put a breakpoint in the process function (maybe just because FireBug was open?).

Anyway, as a patch I re-downloaded Bootstrap with typeahead omitted, got the typeahead from here: https://gist.github./2712048

And used this code:

$(document).ready(function() {
    $('input[name=artist]').typeahead({
        'source': function (typeahead) {
            return $.get('/7d/search-artist.php', { 'artist': typeahead.query }, function (data) {
                return typeahead.process(data);
            });
        },
        'items': 3,
        'minLength': 3
    },'json')
});

My server returns this (for 'Bo'):

["Bo","Bo Burnham","Bo Diddley","Bo Bruce","Bo Carter",
 "Eddie Bo","Bo Bice","Bo Kaspers Orkester","Bo Saris","Bo Ningen"]

Of course, now it ignores my minLength, but it will get me through the day. Hope this helps.

EDIT: Found the solution here: Bootstrap 2.2 Typeahead Issue

Using the typeahead included with Bootstrap 2.2.1, the code should read:

    $(document).ready(function() {
        $('input[name=artist]').typeahead({
            'source': function (query,typeahead) {
                return $.get('/search-artist.php', { 'artist': encodeURIComponent(query) }, function (data) {
                    return typeahead(data);
                });
            },
            'items': 3,
            'minLength': 3
        },'json')
    });

Here's what I do to facilitate remote data sources with bootstrap's typeahead:

$("#search").typeahead({

    source: function(typeahead, query) {

        $.ajax({

            url: "<?php echo base_url();?>customers/search/"+query,
            success: function(data) {

                typeahead.process(data);
            },
            dataType: "json"
        });
    },
    onselect: function(item) {

        $("#someID").val(item.id);
    }
});

And then you just need to make sure your JSON-encoded arrays contain a value index for the label and an id field to set your hidden id afterwards, so like:

$this->load->model('car_model', 'cars');
$brands = $this->cars->searchBrand($this->uri->segment(4));
$output = array();
foreach($brands->result() as $r) {

    $item['value'] = $r->brandName;
    $item['id']    = $r->brandID;
    $output[] = $item;
}
echo json_encode($output);
exit;

$.post is asynchronous, so you can't user return in it. That doesn't return anything.

PHP, returns a JSON encoded array

$this->load->model('car_model', 'cars');
$result = $this->cars->searchBrand($this->input->post('query'));
$this->output->set_status_header(200);
$this->output->set_header('Content-type: application/json');
$output = array();
foreach($result as $r)
    $output['options'][$r->brandID] = $r->brandName;
print json_encode($output);

Outputs: {"options":{"9":"Audi","10":"Austin","11":"Austin Healey"}}

JS updated:

$(".searchcarBrands").typeahead({
    source: function(query, typeahead) {
        $.ajax({
            url: site_url + '/cars/search_brand/'+query,
            success: function(data) {
                typeahead.process(data);
            },
            dataType: "json"
        });
    },
    onselect: function(item) {
        $("#someID").val(item.id);
    }
});

UPDATE: Uncaught TypeError: Object function (){return a.apply(c,e.concat(k.call(arguments)))} has no method 'process'

If I type just 'A' then typeahead shows me only the first letter of each result (a bunch of A letters). If I type a second letter I see nothing anymore.

I've tried JSON.parse on the data or using data.options but no luck.

What am I doing wrong?

PHP, returns a JSON encoded array

$this->load->model('car_model', 'cars');
$result = $this->cars->searchBrand($this->input->post('query'));
$this->output->set_status_header(200);
$this->output->set_header('Content-type: application/json');
$output = array();
foreach($result as $r)
    $output['options'][$r->brandID] = $r->brandName;
print json_encode($output);

Outputs: {"options":{"9":"Audi","10":"Austin","11":"Austin Healey"}}

JS updated:

$(".searchcarBrands").typeahead({
    source: function(query, typeahead) {
        $.ajax({
            url: site_url + '/cars/search_brand/'+query,
            success: function(data) {
                typeahead.process(data);
            },
            dataType: "json"
        });
    },
    onselect: function(item) {
        $("#someID").val(item.id);
    }
});

UPDATE: Uncaught TypeError: Object function (){return a.apply(c,e.concat(k.call(arguments)))} has no method 'process'

If I type just 'A' then typeahead shows me only the first letter of each result (a bunch of A letters). If I type a second letter I see nothing anymore.

I've tried JSON.parse on the data or using data.options but no luck.

What am I doing wrong?

• javascript
• twitter-bootstrap

Share Improve this question Follow edited Nov 6, 2012 at 15:32 stef asked Nov 6, 2012 at 13:48 stefstef 27.8k3131 gold badges107107 silver badges143143 bronze badges 2

• Are you essentially attempting to convert typeahead to allow a remote data source? – Darrrrrren Commented Nov 6, 2012 at 13:55
• OK, see my answer - I believe my code should help you. – Darrrrrren Commented Nov 6, 2012 at 14:00

Add a ment  | 

3 Answers 3

Sorted by: Reset to default 3

I've been battling this for the last day with Bootstrap 2.2.1. No matter what I did, it would not work. For me, I always got the process undefined error unless I put a breakpoint in the process function (maybe just because FireBug was open?).

Anyway, as a patch I re-downloaded Bootstrap with typeahead omitted, got the typeahead from here: https://gist.github./2712048

And used this code:

$(document).ready(function() {
    $('input[name=artist]').typeahead({
        'source': function (typeahead) {
            return $.get('/7d/search-artist.php', { 'artist': typeahead.query }, function (data) {
                return typeahead.process(data);
            });
        },
        'items': 3,
        'minLength': 3
    },'json')
});

My server returns this (for 'Bo'):

["Bo","Bo Burnham","Bo Diddley","Bo Bruce","Bo Carter",
 "Eddie Bo","Bo Bice","Bo Kaspers Orkester","Bo Saris","Bo Ningen"]

Of course, now it ignores my minLength, but it will get me through the day. Hope this helps.

EDIT: Found the solution here: Bootstrap 2.2 Typeahead Issue

Using the typeahead included with Bootstrap 2.2.1, the code should read:

    $(document).ready(function() {
        $('input[name=artist]').typeahead({
            'source': function (query,typeahead) {
                return $.get('/search-artist.php', { 'artist': encodeURIComponent(query) }, function (data) {
                    return typeahead(data);
                });
            },
            'items': 3,
            'minLength': 3
        },'json')
    });

Here's what I do to facilitate remote data sources with bootstrap's typeahead:

$("#search").typeahead({

    source: function(typeahead, query) {

        $.ajax({

            url: "<?php echo base_url();?>customers/search/"+query,
            success: function(data) {

                typeahead.process(data);
            },
            dataType: "json"
        });
    },
    onselect: function(item) {

        $("#someID").val(item.id);
    }
});

And then you just need to make sure your JSON-encoded arrays contain a value index for the label and an id field to set your hidden id afterwards, so like:

$this->load->model('car_model', 'cars');
$brands = $this->cars->searchBrand($this->uri->segment(4));
$output = array();
foreach($brands->result() as $r) {

    $item['value'] = $r->brandName;
    $item['id']    = $r->brandID;
    $output[] = $item;
}
echo json_encode($output);
exit;

$.post is asynchronous, so you can't user return in it. That doesn't return anything.

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