[英]I am trying to update records by php code. but having trouble in my code
I am trying to update records by php code. 我正在尝试通过php代码更新记录。 but having trouble in my code.
但是我的代码有麻烦。
code is running without using functions (get40(),show()) but i want to use buttons to call these functions. 代码运行时未使用函数(get40(),show()),但我想使用按钮来调用这些函数。 code is give below:
代码如下:
this is main code file: 这是主代码文件:
<?php
$data;
$count=0;
$host='localhost';
$uname='root';
$pass="";
$dbname="testing";
$db=new mysqli($host,$uname,$pass,$dbname);
$result;
$query;
if (mysqli_connect_errno()) {
echo 'Error: Could not connect to database. Please try again later.';
exit;
}
function get40($off='0', $rcd='2')
{
global $db,$result,$query;
$query="SELECT sword, tword, reviewed from TempDictionary LIMIT $off,$rcd";
$result = $db->query($query)or trigger_error($db->error);
}
function show()
{
global $db,$result,$query;
echo "<form action='testSave.php' method='POST'>";
echo "<table border=1 >";
while($row = mysqli_fetch_array($result))
{
global $count;
$count+=1;
echo "<tr>";
echo "<td><input type='text' name='s[]' value=" . $row['sword'] . "></td>";
echo "<td><input type='text' name='t[]' value=" . $row['tword'] . "></td>";
echo "<td><input type='text' name='r[]' size='1' value=" .$row['reviewed'] . "></td>";
echo "</tr>";
}
echo "</table>";
echo "<input type='text' name='no' size='2' value=" .$count. ">";
echo "<input type='submit' value='Save'>";
echo "</form>";
}
//get40(1,2);
//show();
mysqli_close($db);
?>
<script type="text/javascript">
function get()
{
<?php
get40(0,3);
show();
?>
}
</script>
<form action="" method="POST">
<input type=button name=btnGet value=Get onclick="get()">
</form>
and savetest.php file is: 而savetest.php文件是:
<?php
$no=$_POST['no'];
$i=0;
$s=$_POST['s'];
$t=$_POST['t'];
$r=$_POST['r'];
$host='localhost';
$uname='root';
$pass="";
$dbname="testing";
$db=new mysqli($host,$uname,$pass,$dbname);
if (mysqli_connect_errno()) {
echo 'Error: Could not connect to database. Please try again later.';
exit;
}
// display
foreach ($s as $key=>$value)
{
print "$key = $value $t[$key] $r[$key]";
echo '<br>';
}
// save to DB
foreach ($s as $key=>$value)
{
// $query="UPDATE TempDictionary set sword=".$value.",tword=".$t[$key].",reviewed=".$r[$key]."
// WHERE sword=".$value;
$query="UPDATE TempDictionary set sword='$value',tword='$t[$key]',reviewed='$r[$key]'
WHERE sword='$value'";
$result = $db->query($query)or trigger_error($db->error);
}
mysqli_close($db);
?>
the problem is while i click button it does not works. 问题是当我单击按钮时,它不起作用。
AFAIK you are trying to run PHP code with Javascript code. AFAIK,您正在尝试使用Javascript代码运行PHP代码。 Use Jquery/Ajax to solve your problem.
使用Jquery / Ajax解决您的问题。 You can Google for that or look at stackoverflow.
您可以为此搜索Google或查看stackoverflow。
Step 1: Learn how a click on a button can run PHP Run php function inside jQuery click 第1步:了解单击按钮如何运行PHP,并在jQuery click中运行php函数
Step 2: Learn how to use data from a database and populate a form with it: ajax call to populate form fields from database query when select value changes 第2步:了解如何使用数据库中的数据并用它填充表单 : 选择值更改时,ajax调用从数据库查询中填充表单字段
That's the first step in solving your problem. 这是解决问题的第一步。 (sorry it's no copy/paste solution, just a pointer in the right direction)
(很抱歉,这不是复制/粘贴的解决方案,只是指向正确方向的指针)
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