无法使用PHP查询将图像从一个表导入另一个表 [英] Failed to import image from one table to another using PHP query
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问题描述
事情是图像已经在一个表中并保存在文件夹中，
但是我想把它放在另一个具有相同属性的表上。
无论如何都可以吗？
这是编辑表1中数据的代码：
the thing is the image is already in one table and saved in folder,
but i want to put it on another table with same properties.
is there anyway to do it?
this is the code to edit data from table 1 :
$aksi="modul/mod_berita/aksi_berita2.php";
case "editberita":
$edit = mysqli_query($koneksi,"SELECT * FROM berita_sementara WHERE id_berita='$_GET[id]'");
$r = mysqli_fetch_array($edit);
echo "<h2>Edit Berita</h2>
<form method=POST enctype='multipart/form-data' action=$aksi?rajaampat=berita&act=kirim>
<input type=hidden name=id value=$r[id_berita]>
<table>
<tr><td width=70>Judul</td> <td> : <input type=text name='judul' size=60 value='$r[judul]'></td></tr>
<tr><td>Kategori</td> <td> : <select name='kategori'>";
$tampil=mysqli_query($koneksi,"SELECT * FROM kategori ORDER BY nama_kategori");
if ($r[id_kategori]==0){
echo "<option value=0 selected>- Pilih Kategori -</option>";
}
while($w=mysqli_fetch_array($tampil)){
if ($r[id_kategori]==$w[id_kategori]){
echo "<option value=$w[id_kategori] selected>$w[nama_kategori]</option>";
}
else{
echo "<option value=$w[id_kategori]>$w[nama_kategori]</option>";
}
}
echo "</select></td></tr>";
if ($r[headline]=='Y'){
echo "<tr><td>Headline</td> <td> : <input type=radio name='headline' value='Y' checked>Y
<input type=radio name='headline' value='N'> N</td></tr>";
}
else{
echo "<tr><td>Headline</td> <td> : <input type=radio name='headline' value='Y'>Y
<input type=radio name='headline' value='N' checked>N</td></tr>";
}
echo "<tr><td>Isi Berita</td> <td> <textarea name='isi_berita' style='width: 600px; height: 350px;'>$r[isi_berita]</textarea></td></tr>
<tr><td>Gambar</td> <td> : ";
if ($r['gambar']!=''){
echo "<img src='../foto_berita/small_$r[gambar]' name='fupload'>";
}
echo "</td></tr>
<tr><td>Ganti Gbr</td> <td> : <input type=file name='fupload' size=30> *)</td></tr>
<tr><td colspan=2>*) Apabila gambar tidak diubah, dikosongkan saja.</td></tr>";
$d = GetCheckboxes('tag', 'tag_seo', 'nama_tag', $r[tag]);
echo "<tr><td>Tag (Label)</td><td> $d </td></tr>";
echo "<tr><td colspan=2><input type=submit value=Post>
<input type=button value=Batal onclick=self.history.back()></td></tr>
</table></form>";
break;
这是用于将表1中已经编辑过的数据插入表2中的代码： br $>
and this is the code for inserting the data that already been edited in table 1 into table 2 :
elseif ($rajaampat=='berita' AND $act=='kirim'){
$lokasi_file = $_FILES['fupload']['tmp_name'];
$tipe_file = $_FILES['fupload']['type'];
$nama_file = $_FILES['fupload']['name'];
$acak = rand(1,99);
$nama_file_unik = $acak.$nama_file;
if (!empty($_POST['tag_seo'])){
$tag_seo = $_POST['tag_seo'];
$tag=implode(',',$tag_seo);
}
$judul_seo = seo_title($_POST['judul']);
// Apabila ada gambar yang diupload
if (!empty($lokasi_file)){
if ($tipe_file != "image/jpeg" AND $tipe_file != "image/pjpeg"){
echo "<script>window.alert('Upload Gagal, Pastikan File yang di Upload bertipe *.JPG');
window.location=('../../media.php?rajaampat=berita)</script>";
}
else{
UploadImage($nama_file_unik);
mysqli_query($koneksi,"INSERT INTO berita(judul,
judul_seo,
id_kategori,
headline,
username,
isi_berita,
jam,
tanggal,
hari,
tag,
gambar)
VALUES('$_POST[judul]',
'$judul_seo',
'$_POST[kategori]',
'$_POST[headline]',
'$_SESSION[namauser]',
'$_POST[isi_berita]',
'$jam_sekarang',
'$tgl_sekarang',
'$hari_ini',
'$tag',
'$nama_file_unik')");
header('location:../../media.php?rajaampat='.$rajaampat);
}
}
else{
mysqli_query($koneksi,"INSERT INTO berita(judul,
judul_seo,
id_kategori,
headline,
username,
isi_berita,
jam,
tanggal,
tag,
hari)
VALUES('$_POST[judul]',
'$judul_seo',
'$_POST[kategori]',
'$_POST[headline]',
'$_SESSION[namauser]',
'$_POST[isi_berita]',
'$jam_sekarang',
'$tgl_sekarang',
'$tag',
'$hari_ini')");
header('location:../../media.php?rajaampat='.$rajaampat);
}
$jml=count($tag_seo);
for($i=0;$i<$jml;$i++){
mysqli_query($koneksi,"UPDATE tag SET count=count+1 WHERE tag_seo='$tag_seo[$i]'");
}
}
我的尝试：
所有数据都很好，可以插入新表，但图像不是，它是空的。
任何解决方案都很受欢迎
What I have tried:
all the data is fine and can be inserted to the new table, but the image is not, it is empty.
any solution is appreciated
推荐答案
aksi =modul / mod_berita / aksi_berita2.php;
caseeditberita：
aksi="modul/mod_berita/aksi_berita2.php"; case "editberita":
edit = mysqli_query（
edit = mysqli_query(
koneksi，SELECT * FROM berita_sementara WHERE id_berita ='
koneksi,"SELECT * FROM berita_sementara WHERE id_berita='
这篇关于无法使用PHP查询将图像从一个表导入另一个表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！
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