SELECT always returns the last record

Question

SELECT always returns the last record

Navigation

#1 by (2 votes)

0

I'm trying to make the moment that I select an option in a select form, other fields are filled through this information. I've been able to do this partially, the last record in the table I'm looking for is always selected, but it actually needs the selected record in option .

Another problem is the mask that I make for the value does not hold when I change the value, I do not know how to use a mask that always stays while editing the value of the field.

If someone knows what I'm doing wrong, thank you if you can help!

<div class="form-group">
    <label for="exampleInputPassword1">Descrição</label>
    <select class="form-control m-b-10" required="required" name="pan_id" charset="utf-8" onchange="adiciona()">
        <option>Selecione a Descrição</option>
        <?php
            $select  = "SELECT * FROM p";
            $res  = mysqli_query($conexao, $select);
            while($mostra = mysqli_fetch_assoc($res)){
                $id     = $mostra['id'];
                echo '<option  charset="utf-8" value = '. $id . '>' . $mostra['descricao'] . '</option>';
            }
        ?>
    </select>
</div>

<?php 
    $select  = "SELECT  local, valor, tipo_valor FROM p WHERE id = '$id'"; 
    $resul  = mysqli_query($conexao, $select);

    while($mostrar = mysqli_fetch_assoc($resul)){
        $pan_valor   = $mostrar['valor'];
        $pan_local   = $mostrar['local'];
        $pan_tipo_valor = $mostrar['tipo_valor'];
    }                                    
?> 

<script>
    function adiciona(){
        var valor       = '<?php echo 'R$' . number_format($valor, 2, ',', '.');?>';
        var local       = '<?php echo $local; ?>';
        var vtipo_valor = '<?php echo $tipo_valor ?>';

        document.getElementById('valor').value = ''; 
        document.c.valor.value += valor;

        document.getElementById('local').value = '';
        document.c.local.value += local;

        switch(vtipo_valor) {
            case 'D':
            document.getElementById("tipo_valor").innerHTML = "<option selected>DIA</option><option>MÊS</option><option>ANO</option>";
            break;

            case 'M':
            document.getElementById("tipo_valor").innerHTML = "<option>DIA</option><option selected>MÊS</option><option>ANO</option>";
            break;

            case 'A':
            document.getElementById("tipo_valor").innerHTML = "<option>DIA</option><option>MÊS</option><option selected>ANO</option>";
        }
</script>

I'm trying to do this:

      function adiciona(){
          var xhttp = new XMLHttpRequest();
          var str   = document.getElementById('descricao');
          var desc  = str.value;
          xhttp.onreadystatechange = function() {
            if (this.readyState == 4 && this.status == 200) {
              document.getElementById("local").innerHTML = this.responseText;
              xhttp.open("GET", "php/busca.php?desc=" + desc, true);
              xhttp.send();
            }
          };

      }

No search.php:

<?php
  $id      = $_GET['desc']; 
  $conexao = mysqli_connect('...', '', '', '');
  if(!$conexao){
    echo "<script> window.location.replace('../erro.html'); </script>"; 
}
$select  = "SELECT  local, valor, tipo_valor FROM p WHERE id = '$id'";
$resul  = mysqli_query($conexao, $select);
while($mostrar = mysqli_fetch_assoc($resul)){
    $valor      = $mostrar['valor'];
    $local      = $mostrar['local'];
    $tipo_valor = $mostrar['tipo_valor'];
}

php javascript html mysqli

asked by anonymous 26.04.2017 / 21:29

1 answer

Results of Football Games on the site How to get current url with PHP

score 2 · Accepted Answer

Exactly the expected behavior is happening. For you to achieve the dynamic selection, or you make a ajax request to each selection of an option of your select (to be done a specific search by id), or you return the entire contents of the variable $ result as vector (not feasible, due to exponential growth over time) to be manipulated by javascript.

Example

You can make an ajax request using api fetch . Here's a small example that suits your needs.

Your current file:

<div class="form-group">
    <label for="exampleInputPassword1">Descrição</label>
    <select class="form-control m-b-10" required="required" name="pan_id" charset="utf-8" onchange="adiciona();">
        <option>Selecione a Descrição</option>
        <option  charset="utf-8" value ="1" >Primeiro id</option>;
        <option  charset="utf-8" value ="2" >Segundo id</option>;
        <option  charset="utf-8" value ="3" >Terceiro id</option>;    
    </select>
</div>

<!-- Coloque sua consulta no arquivo ajax.php-->

<!-- Campo para testar -->
<select id="tipo_valor"></select>

<script>
    //agora a funcao adiciona faz uma requisição ajax usando a api fetch
    function adiciona(){
        //obtem o valor do select com name igual a pan_id
        var id_selecionado = document.getElementsByName("pan_id")[0].value;

        var parametros = {
                method: 'GET'
        };

        //definie a url a ser acessada, e executa um callback na resposta da requisição
        fetch('ajax.php?id=' + id_selecionado, parametros).then(function(resposta) {
            return resposta.json();
        }).then(function(retorno){
            console.log(retorno);
            switch(retorno.tipo_valor) {
            case 'D':
            document.getElementById("tipo_valor").innerHTML = "<option selected>DIA</option><option>MÊS</option><option>ANO</option>";
            break;

            case 'M':
            document.getElementById("tipo_valor").innerHTML = "<option>DIA</option><option selected>MÊS</option><option>ANO</option>";
            break;

            case 'A':
            document.getElementById("tipo_valor").innerHTML = "<option>DIA</option><option>MÊS</option><option selected>ANO</option>";

            }
        });
    }
</script>

Note: there have been some changes in html to make it easier to debug, undo.

And create a new php file in the same directory as the previous one.

ajax.php

<?php
header('Content-Type: json/application');
$id = $_GET['id'];

$select  = "SELECT  local, valor, tipo_valor FROM p WHERE id = '$id'"; 
$resul  = mysqli_query($conexao, $select);

//supondo que na sua tabela p  tenha o campo tipo_valor
$mostrar = mysqli_fetch_assoc($resul);

//desde que mostrar tenha $mostrar['tipo_valor'];
echo json_encode($mostrar);