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php - 如何使用php将记录插入到两个mysql表中?

转载 作者:行者123 更新时间:2023-11-29 01:39:18 26 4
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我的数据库中有两个表,一个是“test1”,第二个是“test2”。测试 1 的字段是“id”,“survey_no”,“lastname”,对于测试字段,我有“id”,“survey_no”,“address”,test1 和 test2 的 id 是主键并设置为自动-增量..我在这里想要的是我在test1中插入“survey_no”的内容也应该插入到“test2中的survey_no”中..有人可以帮我吗..请..

我这里有插入代码 sample.php

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head>

<body>

<form method="POST" class="signin" action="" name="add" id="form1">
<fieldset class="textbox">
<label class="province_id">
<span>province id</span>
<input id="province" name="survey_no" type="number" value="" autocomplete="on" placeholder="survey_no">
</label>
<label class="municipality">
<span>Municipality</span>
<input id="municipality" name="lastname" value="" type="text" autocomplete="on" placeholder="lastname">
</label>
<br />
<br />
<button id="submit" type="submit" name="submit">Save</button>
<button id="submit" type="reset" name="reset">Reset</button>
</fieldset>


<?php
$conn=mysql_connect('localhost','root','');
if(!$conn)
{
die('could not connect:' .mysql_error());
}
mysql_select_db("sample",$conn);

if(isset($_POST['submit']))
{

$survey_no=$_POST['survey_no'];
$lastname=$_POST['lastname'];

$result = mysql_query("INSERT INTO `test1`(survey_no,lastname)
VALUES ('$survey_no','$lastname')");

if($result)
{
echo "<script type=\"text/javascript\">".
"alert('Saved!');".
"</script>";
}
else{
echo "<script type=\"text/javascript\">".
"alert('Failed!');".
"</script>";
}
}

?>

</form>


</body>
</html>

最佳答案

到目前为止我知道你想要这个

$result = mysql_query("INSERT INTO `test1`(survey_no,lastname) 
VALUES ('$survey_no','$lastname')");


$result1 = mysql_query("INSERT INTO `test2`(survey_no)
VALUES ('$survey_no')");

关于php - 如何使用php将记录插入到两个mysql表中?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/30228944/

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