从sql中获取信息并将其放入表单中 [英] Get information out of sql and put it in a form

问题描述

所以我需要从sql中获取信息并将其放在我的表单中的下拉列表中。这就是我所拥有的......我非常迷茫......信息已被预填充到sql中。我认为最重要的部分是相对正确的，然后我不知道如何在表格中引用它。

PHP

 <？php 
 $ id = $ _GET ['id']; 
 
 $ conn = mysql_connect（localhost，root，）或die（mysql_error（））; 
 
 mysql_select_db（赋值3，$ conn）; 
 
 $ sql =select schoolname FROM schooltable WHERE id = $ id; 
 
 $ result = mysql_query（$ sql，$ conn）或die（mysql_error（））; 
 
（$ row = mysql_fetch_assoc（$ result））{
 foreach（$ row as $ name => $ value）{
 print$ name = $ value< / BR>中; 
} 
} 
 
 mysql_data_seek（$ result，0）; 
 while（$ row = mysql_fetch_assoc（$ result））{
 //从用户列表中选择id，firstname，lastname 
 $ school = $ row [schoolname]; 
 $ grad = $ row [lastname]; 
} 
？> 
  

HTML

 < div class =form-group> 
< label class ='col-xs-4 control-label'>你去过什么学校为本科学习？ < /标签> 
< div class ='col-xs-8'> 
< select class =form-control backgroundid ='dropdown'> 
< option><？php print $ schoolname？>< / option> 
< option><？php print $ schoolname？>< / option> 
< option><？php print $ schoolname？>< / option> 
< option><？php print $ schoolname？>< / option> 
< option value =bing><？php print $ schoolname？>< / option> 
< / select> 
< input type =hiddenname =idid ='id'value =<？php print $ id？>> 
< input type =hiddenname =editModevalue =edit> 
< / div> 
 

解决方案

 <？php 
 if（！isset（$ _ GET ['id']]）{
 $ echo'id = not in present in URL。Exiting。'; 
 return false; 
} 
 $ id = intval（$ _ GET ['id']）; 
 
 $ conn = mysql_connect（localhost，root，MIS42520！$）或die（mysql_error（））; 
 $ b $ mysql_select_db（assignment 3，$ conn）; 
 
 $ sql =select * FROM schooltable WHERE id ='。mysql_real_escape_string（$ id）。'; 
 
 $ result = mysql_query（$ sql，$ conn）或die（mysql_error（））; 
 
 $ schools = array（）; 
 while（$ row = mysql_fetch_assoc（$ result））{
 $ [] = $ row; 
 
 $> 
 
< div class =form-group> 
< label class ='col-xs-4 control-label '>你去学校的什么学校？< / label> 
< div class ='col-xs-8'> 
< select class =form-control background id = 'dropdown'> 
<？php foreach（$ schools as $ school）{？> 
< option value =<？php echo $ school ['schoolname'];？>><？php echo $ school ['schoolname'];？>< / option> 
<？php}？> 
< / select> 
< input type =hiddenname =idid ='id'value =<？php echo $ id？>> 
< input type =hiddenname =editModevalue =edit> 
< / div> 
< / div 
  

So I need to get information out of sql and put it in my dropdown in my form. Here's what I have... I'm very lost.. The info has been prepopulated into sql. I believe the top part is relatively right and then I don't know how to reference it in the form.

PHP

<?php
$id= $_GET['id'];

$conn = mysql_connect("localhost", "root", "") or die (mysql_error());

mysql_select_db("assignment 3", $conn);

$sql = "select schoolname FROM schooltable WHERE id=$id";

$result=mysql_query($sql, $conn) or die(mysql_error());

while ($row=mysql_fetch_assoc($result)){
    foreach($row as $name => $value){
        print "$name = $value</br>";
    }
}

mysql_data_seek($result, 0);
while ($row=mysql_fetch_assoc($result)){
    //select id, firstname, lastname from userlist
    $school = $row["schoolname"];
    $grad = $row["lastname"];
}
?>

HTML

                     <div class="form-group">
                    <label class='col-xs-4 control-label'>What school did you go to for your undergrad?  </label>
                    <div class='col-xs-8'>
                        <select class="form-control background" id='dropdown'>
                          <option>"<?php print $schoolname ?>"</option>
                          <option>"<?php print $schoolname ?>"</option>
                          <option>"<?php print $schoolname ?>"</option>
                          <option>"<?php print $schoolname ?>"</option>
                          <option value="bing">"<?php print $schoolname ?>"</option>
                        </select>
                        <input type="hidden" name="id" id='id' value="<?php print $id ?>">
                        <input type="hidden" name="editMode" value="edit">
                    </div>
                </div

解决方案

Here you go

<?php
if(!isset($_GET['id']]){
    echo 'id= not present in URL. Exiting.';
    return false;
}
$id = intval($_GET['id']);

$conn = mysql_connect("localhost", "root", "MIS42520!$") or die (mysql_error());

mysql_select_db("assignment 3", $conn);

$sql = "select * FROM schooltable WHERE id='" . mysql_real_escape_string($id) . "'";

$result = mysql_query($sql, $conn) or die(mysql_error());

$schools = array();
while ($row = mysql_fetch_assoc($result)) {
    $schools[] = $row;
}
?>

<div class="form-group">
    <label class='col-xs-4 control-label'>What school did you go to for your undergrad?  </label>
    <div class='col-xs-8'>
        <select class="form-control background" id='dropdown'>
            <?php foreach($schools as $school){?>
                <option value="<?php echo $school['schoolname'];?>"><?php echo $school['schoolname'];?></option>
            <?php } ?>
        </select>
        <input type="hidden" name="id" id='id' value="<?php echo $id ?>">
        <input type="hidden" name="editMode" value="edit">
    </div>
</div

这篇关于从sql中获取信息并将其放入表单中的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！