带有“其他"标记的html单选按钮具有文本框的选择 [英] html radio button with an "other" selection that has a text box

问题描述

当我的插入程序运行时，它将在文本框中写入文本，但是如果我选择单选按钮而不是其他"，则不会在表中插入任何内容. 如果我删除其他"代码，则选中的单选按钮将被写入表格.

When my insert runs it will write the text in the text box but if I choose a radio button selection other than "other" nothing is inserted into the table. If I remove the "other"code the radio button selected is written to the table.

这是代码

<input type="radio" name="job" value="PHP Programmer">PHP Programmer
<input type="radio" name="job" value="SQL Programmer">SQL Programmer<br>
<input type="radio" name="job">Other <input type="text" name="job" >

<h3>* If yes, check which ISO standard(s) you are accredited?</h3>
<input type="radio" name="iso_standard" value="17020">17020
<input type="radio" name="iso_standard" value="17025">17025

<br />
<input action="submit" type="submit" name="submit" value="Submit">
</form>

<?php
  if(isset($_POST['job']) && isset($_POST['iso_standard']))
  { 
$job = mysqli_real_escape_string($db, $_POST['job']);
$iso_standard =mysqli_real_escape_string($db, $_POST['iso_standard']);


  $sql="insert into tbl_test_insert(iso_cert, iso_standard)
  values ('$job', '$iso_standard')";

  if(!mysqli_query($db, $sql))
  {
die('Error: ' .mysqli_error($db));
  }
  echo "1 record added";
  }
  else
  {
echo "You didn't choose all the options!
  }

  ?>

推荐答案

您正在用文本输入字段覆盖job的值.不管有没有值，它都不会计算.

You are overwriting the value of job with the text input field. It does not mather if it has any value or not.

请将您的<input type="text" name="job" >重命名为例如otherJob，您应该没事

Please rename your <input type="text" name="job" > to for example otherJob and you should be fine

更改<input type="radio" name="job">Other <input type="text" name="job" >

到<input type="radio" name="job" value="other">Other <input type="text" name="otherJob">

更改$job = mysqli_real_escape_string($db, $_POST['job']);

到

  $job = mysqli_real_escape_string($db, $_POST['job']);
  $otherJob = mysqli_real_escape_string($db, $_POST['otherJob']);
  if ($job == 'other') {
     $jobField = $otherJob;
  } else {
     $jobField = $job;
  }

并更改$sql="insert into tbl_test_insert(iso_cert, iso_standard) values ('$job', '$iso_standard')";

and change $sql="insert into tbl_test_insert(iso_cert, iso_standard) values ('$job', '$iso_standard')";

到$sql="insert into tbl_test_insert(iso_cert, iso_standard) values ('$jobField', '$iso_standard')";

to $sql="insert into tbl_test_insert(iso_cert, iso_standard) values ('$jobField', '$iso_standard')";

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