嗨我是一个初学者在php和mysql这里如果有问题,它是我的isset帖子提交不工作或我的编码有一些错误,谁能告诉我在编码中我错过了什么?提前谢谢你:)
这是我的php:
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "tempahperalatan";
// Create connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
if (isset($_POST['submit'])) {
$pemohon = $_POST['namaPemohon'];
$trkhMula = $_POST['tmula'];
$trkhAkhir = $_POST['takhir'];
$n_program = $_POST['namaProgram'];
$lokasi = $_POST['lokasi'];
$n_anjuran = $_POST['namaAnjuran'];
$catatan = $_POST['catatan'];
$sql = "INSERT INTO daftartempah (pemohon, trkhMula, trkhAkhir, n_program, lokasi, n_anjuran, catatan) VALUES ('$namaPemohon', '$tmula', '$takhir', '$namaprogram', '$lokasi', '$namaAnjuran', '$catatan')";
if (mysqli_query($conn, $sql)) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . mysqli_error($conn);
}
}
mysqli_close($conn);
?>
这是我的表格:
<form action="page1.php" method="POST">
<div class="form-group row text-left">
<label for="example-text-input" class="col-2 col-form-label">Nama Pemohon: </label>
<div class="col-10">
<input class="form-control" type="text" name="namaPemohon" id="namaPemohon">
</div>
</div>
<div class="form-group row text-left">
<label for="example-date-input" class="col-2 col-form-label">Tarikh Mula: </label>
<div class="col-10">
<input class="form-control" type="date" value="0000-00-00" name="tmula" id="tmula">
</div>
</div>
<div class="form-group row text-left">
<label for="example-date-input" class="col-2 col-form-label">Tarikh Akhir: </label>
<div class="col-10">
<input class="form-control" type="date" value="0000-00-00" name="takhir" id="takhir">
</div>
</div>
<div class="form-group row text-left">
<label for="example-text-input" class="col-2 col-form-label">Nama Program: </label>
<div class="col-10">
<input class="form-control" type="text" name="namaProgram" id="namaProgram">
</div>
</div>
<div class="form-group row text-left">
<label for="example-text-input" class="col-2 col-form-label">Lokasi: </label>
<div class="col-10">
<input class="form-control" type="text" name="lokasi" id="lokasi">
</div>
</div>
<div class="form-group row text-left">
<label for="example-text-input" class="col-2 col-form-label">Dept/Kelab/Anjuran: </label>
<div class="col-10">
<select class="form-control" name="namaAnjuran" id="namaAnjuran">
<option name="1" value="1">BK - B. Kewangan</option>
<option name="2" value="1">BP - B. Pentadbiran</option>
<option name="3" value="1">BPK - B. Perancangan Hal Ehwal Korporat</option>
<option name="4" value="1">BPPP - B. Pemb. Penilaian Projek</option>
<option name="5" value="1">BPPS - B. Pemb. Projek Sosial</option>
<option name="6" value="1">UAD - Unit Audit Dalaman</option>
<option name="7" value="1">PWT - Pej. wilayah Temerloh</option>
<option name="8" value="1">PWB - Pej. Wilayah Barat</option>
<option name="9" value="1">KOOP - Koperasi LKPP</option>
<option name="10" value="1">KSKLKPP - KSK LKPP Negeri Pahang</option>
<option name="11" value="1">KSKCIP - KSK LKPP Cawangan Ibu Pejabat</option>
<option name="12" value="1">IBUPEJ - IBU PEJABAT</option>
<option name="13" value="1">KESATUAN - KESATUAN SEKERJA</option>
</select>
</div>
</div>
<div class="form-group row text-left">
<label for="exampleTextarea" class="col-2 col-form-label">Catatan: </label>
<div class="col-10">
<textarea class="form-control" id="exampleTextarea" rows="3" name="catatan" id="catatan"></textarea>
</div>
</div>
<center><button type="submit" class="btn btn-info">Submit</button></center>
</form>
当我点击提交时,没有错误,但数据没有插入数据库。
答案 0 :(得分:1)
试试这样:
<button type="submit" class="btn btn-info" name="submit" value="submit" >Submit</button>
答案 1 :(得分:1)
if (isset($_POST['submit'])) {
这不起作用,原因是您没有提供提交按钮的名称<button type="submit" class="btn btn-info">Submit</button>
添加提交按钮的名称,如下所示。
<button type="submit" name="submit" class="btn btn-info">Submit</button>
OR
if (isset($_POST)) {
答案 2 :(得分:1)
如果要将数据从表单传递到php脚本,则必须记住为该数据分配了name属性,以便在php脚本中查找。
<button type="submit" class="btn btn-info">Submit</button>
在按钮中,您没有分配名称属性。所以在php脚本之后的post方法isset($ _ POST ['submit'])没有找到任何提交属性。所以它返回false。
<button type="submit" class="btn btn-info" name="submit" value="submit" >Submit</button>
所以你必须使用这一行作为按钮。
答案 3 :(得分:1)
始终记住,这是将表格链接到php脚本的name属性。因此应该是:
<button type="submit" name="submit" class="btn btn-info">Submit</button>
答案 4 :(得分:0)
只需检查$_POST
是否为空
if(!empty($_POST)) // form has been submitted
您的案例
中没有名称为submit
的表单输入
答案 5 :(得分:0)
您可以按下按钮的名称
<center><button type="submit" name="submit" class="btn btn-info">Submit</button></center>
这将起作用