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[英]How do I validate a form that doesnt go to another page yet fills out another form on the same page?
[英]Form which retrieves data from mysql using php and fills another form in the same HTML page
我有一个html页面,该页面从一种形式(例如form1)获取数据,并使用php(单击“提交”)将其存储在mysql数据库中。 现在,我正在使用另一种形式,即form2,如果它已经存储在数据库中,则自动填写form1中的输入字段。
因此,我在form2中提供了两个字段,并编写了一个“ match.php”来尝试在mysql数据库中搜索记录。 当我在form2上单击“提交”时,将运行此命令。 我还包含在“ match.php” js脚本中,以更改表格1的值属性。
但是问题是当我在form2的总和上计时时,显示空白页。 如果找到匹配项,我希望填入form1中的文本框!
match.php
$conn = mysqli_connect($servername, $username, $password, $dbname);
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
$a=$_POST['empname'];
$b=$_POST['vid'];
$sql = "SELECT * FROM table1 where name='$a' AND visitorid='vid'";
$result = mysqli_query($conn, $sql);
while($row = mysqli_fetch_assoc($result))
$b=$row['number'];
echo "
<script>
var form1 = document.forms[0];
var form2 = document.forms[2];
var number= form1.elements['n11'];
number.value=".$b."";
echo " </script>";
form1中
<form enctype="multipart/form-data" action="con3.php" method="post">
<table>
<tr rowspan="500" colspan="500">
<td>
<input type='file' name="userfile" onchange="readURL(this);"/>
<img id="blah" src="#" alt="your image" />
</td>
</tr>
<tr colspan="2" rowspan="2">
<td>VISITOR'S ID</td>
<td><input type="textbox" id="t1" value="automatically generated" name="n1" onfocus="this.blur()"/></td>
<tr colspan="2" rowspan="2">
<td>NAME OF THE EMPLOYEE</td><span style="color:red;">*</span>
<td><input type="textbox" id="t4" name="n4" onblur="f4()"/><span style="color:red;">*</span></td>
</tr>
<tr colspan="2" rowspan="2">
<td>ID PROOF</td>
<td>
<select name="n20" onblur="f38()" >
<option>id</option>
<option>1210312117</option>
</select>
</tr>
<tr colspan="2" rowspan="2">
<td>CONTACT NUMBER ON EMERGENCY</td><span style="color:red;">*</span>
<td><input type="textbox" id="t14" name="n11" onblur="f14()" required/><span style="color:red;">*</span></td>
</tr>
<tr colspan="2" rowspan="2">
<td>INTIME</td>
<td><input type="textbox" id="t17" name="n16" onblur="f17()"/></td>
<td><input type="button" value="PICKTIM" id="t18" onclick="document.getElementById('t17').value = f88()"/></td>
</tr>
<tr colspan="2" rowspan="2">
<td>OUTTIME</td>
<td><input type="textbox" id="t100" name="n17" onblur="f100()"/></td>
</tr>
<td><input type="submit" value="SAVE RECORD" id="t25" onclick="f25()"/></td>
</tr>
</table>
窗口2
<form action="match.php" method="post">
<h3> Old employee?</h3>
Name of the emplyee: <input type="text" name="empname" id="id1" /></br>
Visitor id: <input type ="text" name="vid" id="id2"/> <br/>
<input type="submit" value="submit"/>
</form>
将您的SQL数据结果(您获取-> assoc())存储在$ _SESSION对象中,如下所示:
//Before the DOCTYPE and every other line
<?php session_start(); ?>
//In your php code :
$row = mysqli_fetch_assoc($result->empname)
//I don't use this method of PHP for databases, I use PDO, but the result is the same and so I don't know if this is above is really correct
//but you seem to know what to do
$_SESSION["empname"] = $empname;
//And in your form2
Name of the employee : <?php echo $_SESSION["empname"]; ?>
要在“提交”按钮之后显示对象,请使用session_start()在页面上启动会话。 您可以通过echo $ _SESSION [“ object_name”];显示它们。
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