如何在单选按钮中从数据库加载值？

Question

I need to know if you can do this function ... I have a database that has a "boolean" field that accepts 0 or 1 values And I have an HTML page in which there is a radioButton with values Ok or No I must be able to load the boolean value in the radioButton so if in the database the boolean variable is 1 on the HTML page (after executing the SELECT) the radioButton will have OK value if in the database the boolean variable is 0 on the HTML page (after executing the SELECT) the radioButton will have NO value 我需要知道您是否可以执行此功能...我有一个数据库，该数据库的“布尔”字段接受0或1个值，并且我有一个HTML页面，其中有一个单选按钮，其值必须为Ok或No，我必须能够在radioButton中加载布尔值，因此，如果在数据库中HTML页面上的boolean变量为1（执行SELECT之后），则如果数据库中HTML页面上的boolean变量为0（之后，则radioButton将具有OK值）执行SELECT）单选按钮将没有值

this is the code of my attempts : 这是我尝试的代码：

  <?php $servername = "localhost"; $username = "progettocantiere"; $password = ""; $dbname = "my_progettocantiere"; // Create connection $conn = new mysqli($servername, $username, $password, $dbname); // Check connection if ($conn->connect_error) { die("Connection failed: " . $conn->connect_error); } $idAffidataria = $_GET['idAffidataria']; $sql1 = "SELECT * FROM Affidataria WHERE idAffidataria = '{$idAffidataria}'"; $result = $conn->query($sql1); $details = $result->fetch_array(); $savedNomeCantiere = $details["nomeCantiere"]; $savedAddettoSicurezza = $details["addettoSicurezza"]; $savedMailAffidataria = $details["mailAffidataria"]; $savedContrattoDiAppalto = $details["contrattoDiAppalto"]; // <--- BOOLEAN <---- $result1 = $conn->query($sql1); echo($nomeCantiere); ?> <html> <body> <table> <tr> <td colspan="2" bgcolor="#CDECFD" style="font-weight: bold">Cantiere</td> <td colspan="4" bgcolor="#CDECFD"><input type="text" class="form-control" style="width: 100%;" name="cantiereAffidataria" id="cantiereAffidataria" value="<?php echo $savedNomeCantiere; ?>"/>&nbsp;</td> </tr> <tr> <td colspan="2" bgcolor="#CDECFD" style="font-weight: bold">ContrattoDiAppalto</td> <td bgcolor="#B35556"><form action=""> OK <input type="radio" name="contrattoDiAppalto" id="contrattoDiAppalto" value="<?php echo $savedContrattoDiAppalto; ?>" onchange="color(this)" /> <BR> NO <input type="radio" name="contrattoDiAppalto" id="contrattoDiAppalto" value="<?php echo $savedContrattoDiAppalto; ?>" onchange="color(this)" checked/> </form></td> </tr> </table>

in other words I have to be able to change the radioButton according to the data provided by the SELECT on the database 换句话说，我必须能够根据SELECT在数据库上提供的数据来更改radioButton

I tried to give it the value through <?php echo $savedContrattoDiAppalto; ?>" 我试图通过<?php echo $savedContrattoDiAppalto; ?>"给它赋值<?php echo $savedContrattoDiAppalto; ?>" <?php echo $savedContrattoDiAppalto; ?>" but I doubt it's right .. I'm running out of ideas .. can someone help me? <?php echo $savedContrattoDiAppalto; ?>"但我怀疑这是正确的..我的想法已经用完了..有人可以帮助我吗？

Answer 1

Make an variable $checked = 'checked' 使变量$checked = 'checked'

Then make an condition to check if it's 1 or 0 然后进行条件检查是1还是0

$checked = 'checked';
if($savedContrattoDiAppalto != 1){
    $checked = ''; // Default is checked, if it isn't 1 then it's empty and will not check your radio.
}

And apply that variable to your input attributes. 并将该变量应用于您的输入属性。

    <input type="radio" name="contrattoDiAppalto" id="contrattoDiAppalto" 
value="<?php echo $savedContrattoDiAppalto; ?>" onchange="color(this)" $checked/>

Full code should be : 完整代码应为：

<?php
$servername = "localhost";
$username = "progettocantiere";
$password = "";
$dbname = "my_progettocantiere";

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
} 



$idAffidataria = $_GET['idAffidataria'];

$sql1 = "SELECT * FROM Affidataria WHERE idAffidataria = '{$idAffidataria}'";



    $result = $conn->query($sql1);


    $details = $result->fetch_array();




    $savedNomeCantiere = $details["nomeCantiere"];
    $savedAddettoSicurezza = $details["addettoSicurezza"];
    $savedMailAffidataria = $details["mailAffidataria"];
    $savedContrattoDiAppalto = $details["contrattoDiAppalto"];
    $checked = 'checked';             
    if($savedContrattoDiAppalto != true){
        $checked = ''; // if not true then not checked
    }    



$result1 = $conn->query($sql1);

echo($nomeCantiere);

?>


<html>
<body>

<table>
 <tr>

    <td colspan="2" bgcolor="#CDECFD" style="font-weight: bold">Cantiere</td>
    <td colspan="4" bgcolor="#CDECFD"><input type="text" class="form-control" style="width: 100%;" name="cantiereAffidataria" id="cantiereAffidataria" value="<?php echo $savedNomeCantiere; ?>"/>&nbsp;</td>
  </tr>
  <tr>
  <td colspan="2" bgcolor="#CDECFD" style="font-weight: bold">ContrattoDiAppalto</td>
  <td bgcolor="#B35556"><form action="">


       OK <input type="radio" name="contrattoDiAppalto" id="contrattoDiAppalto" value="<?php echo $savedContrattoDiAppalto; ?>" onchange="color(this)"  $checked/> <BR>
        NO  <input type="radio" name="contrattoDiAppalto" id="contrattoDiAppalto" value="<?php echo $savedContrattoDiAppalto; ?>" onchange="color(this)" $checked/>


</form></td>
  </tr>
</table>

如何在单选按钮中从数据库加载值？

问题描述

1 个解决方案

解决方案1
0 2018-10-03 12:55:39

如何在单选按钮中从数据库加载值？

问题描述

1 个解决方案

解决方案1 0 2018-10-03 12:55:39

解决方案1
0 2018-10-03 12:55:39