Sorted list PHP+AJAX

Question

I have been having the following problem since yesterday, putting together an orderly list as I will search for data in the database. I suggested to do with ajax, I did with the help of a colleague of the forum until a certain part. However, in the search with the bank he does not return me anything. Always from Undefined, I believe he’s not looking in the bank.

Follow code in ajax and php

AJAX Code

$('#buscar').on('click', function(){
    $.ajax({
      type: 'GET',
      url: 'busca.php',
      data: {
        id: $('input[name=id]').val()
      },
      success: function(data){
        $('.table tbody').append('<tr><td>'+data.nome+'</td></tr>')
      }
    })
    return false
  })

PHP code

    $
<?php


    $host='localhost:C:/bd_relatorio/clipp.FDB';
    $dbh=ibase_connect($host,'SYSDBA','masterkey');
    $stmt = 'SELECT NOME, ID_CLIENTE FROM TB_CLIENTE';
    $sth = ibase_query($dbh, $stmt);


    $id = $_GET['id'];
    $sql = ibase_query("SELECT NOME FROM TB_CLIENTE WHERE ID_CLIENTE LIKE '%".$id."%'");
    $row = ibase_fetch_row($sql);
    die(json_encode($row));
?>

Answer 1

Whereas your query returns results, in principle your problem is that the return of AJAX interpreted by Javascript is a string and you’re treating as JSON.

To resolve this, add the property dataType='JSON' at the $.ajax:

$.ajax({
    method: "GET",
    dataType: "JSON", //<-- define que o tipo de retorno será tratado como JSON
    // restante do código

or carry out the parse with JSON.parse

var json = JSON.parse(data);
console.log(json.nome);

Any of the solutions should solve your undefined.

Ps.: Your code is vulnerable to SQL Injection.. Further reading

Update

There is another situation to be addressed. When you are using the following code snippet:

ibase_fetch_row($sql);

The return of PHP is an array similar to this

array(1) { 
    ["nome"]=> string(14) "Gabriel Heming" 
}

Next, you are adding to the variable $retorno['NOME']. That is, your array has become this:

array(1) { 
    ["nome"]=> array(1) { 
        ["nome"]=> string(14) "Gabriel Heming" 
    } 
}

Hence, the index NOME is not necessary. You can do exactly like this:

$retorno = ibase_fetch_row($sql);       
echo json_encode($retorno);

Ps. 2: Closing the tag in PHP (?>) it is better to be omitted.

Update 2

Now that you have updated the PHP code, remove the cipher that is at the beginning of PHP and any other characters or whitespace. You must return only the JSON string.

    $ //<-- remova qualquer caracter que esteja antes da abertura de tag do PHP.
<?php

Answer 2

Just use the command SQL Order By:

$sql = ibase_query("SELECT NOME FROM TB_CLIENTE WHERE ID_CLIENTE LIKE '%".$id."%' ORDER BY ID_CLIENTE ASC");

This way select will bring the data in an orderly manner.

Explanation:

Using the command order by you will sort the values of a column ascending(ASC) or descending(DESC).

SELECT "nome_coluna"
FROM "nome_tabela"
[WHERE "condição"]
ORDER BY "nome_coluna" [ASC, DESC];

Example 1:

SELECT Store_Name, Sales, Txn_Date
FROM Store_Information
ORDER BY Sales DESC;

Example 2:

SELECT * FROM Customers
ORDER BY Country ASC, CustomerName DESC;