Question

我将复选框中的值发布到用户个人资料的数据库中。当用户编辑他/她的个人资料时，我希望检查他们之前选择的复选框，以便他们在更新他们的个人资料后不会丢失该信息。我尝试了许多不同的解决方案，但没有运气。

复选框值将表名称members_teachers输入到一个名为focus的列中，并用逗号分隔，例如艺术，数学，舞蹈等。我不确定我离我很近或远远没有完成我的目标但我非常提供您可以提供的任何帮助或建议。非常感谢您提前

我尝试检查值的代码是

<?php
$focusQuery = mysql_query("SELECT focus FROM members_teachers WHERE id = $member_id") or die;

while ($new_row = mysql_fetch_assoc($focusQuery)){

$focusRow = $row['focus'];

$focusValue = explode(',', $focusRow);

foreach($focusValue as $newFocus){

//echo $newFocus;

//echo "<br/>";

$result = mysql_query("SELECT focus FROM members_teachers WHERE focus LIKE '%$focusRow%'") or die;

if(mysql_num_rows($result) > $newFocus){

$checked = 'checked="checked"';

}

else{

$checked = '';

}

}

}
?>

这是我的HTML

<label for="art-focus">Art</label>
                        <input name="focus[]" type="checkbox" value="Art" <?php echo $checked ?>>

<label for="math-focus">Mathematics</label>
                            <input name="focus[]" type="checkbox" value="Mathematics" <?php echo $checked ?>>

<label for="dance-focus">Dance</label>
                            <input name="focus[]" type="checkbox" value="Dance" <?php echo $checked ?>>

Answer 1

<?php
// Create connection
$con=mysqli_connect("hostname","username","pass","dbname");

// Check connection
if (mysqli_connect_errno($con))
  {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
  }
$result = mysqli_query($con,"SELECT focus FROM members_teachers WHERE id = $member_id"); 
while($row = mysqli_fetch_array($result))
  {
    $focus=explode(",",$row['focus']);

?>
<input type="checkbox" name="focus[]" value="Art" <?php if(in_array("Art",$focus)) { ?> checked="checked" <?php } ?> >
<input type="checkbox" name="focus[]" value="Mathematics" <?php if(in_array("Mathematics",$focus)) { ?> checked="checked" <?php } ?> >
<input type="checkbox" name="focus[]" value="Dance" <?php if(in_array("Dance",$focus)) { ?> checked="checked" <?php } ?> >
<?php
}
?>

Answer 2

<?php

 $focusedValues = array();

$focusQuery = mysql_query("SELECT focus FROM members_teachers WHERE id = $member_id") or die;

while ($row = mysql_fetch_assoc($focusQuery)){

    $focusedValues = explode(',', $row['focus']);
}
?>

<label for="art-focus">Art</label>
                        <input name="focus[]" type="checkbox" value="Art" <?php echo in_array('Art', $checked) ?>>

<label for="math-focus">Mathematics</label>
                            <input name="focus[]" type="checkbox" value="Mathematics" <?php echo in_array('Mathematics', $checked) ?>

<label for="dance-focus">Dance</label>
                            <input name="focus[]" type="checkbox" value="Dance" <?php echo in_array('Dance', $checked) ?>

我不知道你为什么第二次SELECT，这是毫无意义的，你已经知道检查了什么，因为它在$focusedValues。此外，在您的代码中，如果未选中任何内容，$checked将为空，否则为checked="checked"。你显然需要为每个输入都有一个变量，不是吗？

Answer 3

<?php  $hobby = $row['hobbies']; 
    $hobbies = explode (' ', $hobby);
?>
<input type="checkbox" name="hobbies[]" value="cricket"  <?php echo in_array('cricket', $hobbies?'checked':'') ?> >cricket
<input type="checkbox" name="hobbies[]" value="singing"  <?php  echo in_array('singing' , $hobbies ?'checked': '')  ; ?> >singing
<input type="checkbox" name="hobbies[]" value="football" <?php  echo in_array('football', $hobbies  ?'checked': '') ; ?> >footballl

检查数据库值的复选框

3 个答案: