Question

我创建了一个简单的详细信息页面，管理员可以在其中查看数据库中的人员...现在每个人都有一个字段简历，其中的链接指向服务器上文件中的简历...人们不会必须发送他们的简历与表格，但仍然链接出现他们的名字...我怎么能改变这一点，所以当他们没有简历时，它会说没有简历，当有人确实添加他们的简历，它将链接它到文件（工作）...

详细信息页面：

find . -name "file*"|while read fname; do
  echo "$fname"
  for line in $(cat "$fname") do
   FS="[_() ]"
   print   $7 "\t" $10 "\t" $11 > $fname
  done
done

我的简历上传（cv）：

<?php

$servername = "localhost";
$username = "root";
$password = "usbw";
$dbname = "persons";

// CREATE A CONNECTION WITH THE DATABASE
// CONNECTIE MAKEN MET DATABASE
$conn = new mysqli($servername, $username, $password, $dbname);
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
} 

// SELECT TABLE NAMES FROM PERSON, ADDRESS AND CV WHERE address_id IS person_address and cv_id IS person_cv
// SELECTEER VAN TABEL PERSON, ADDRESS AND CV WAAR address_id  GELIJK IS AAN person_address EN cv_id AAN person_cv
$sql = "SELECT person_id, person_firstname, person_lastname, 
                 person_email, person_phonenumber,  
                 address_street,address_housenumber, 
                 address_city,address_state,address_zipcode, cv_path
          FROM person 
            inner join address on address.address_id = person.person_address 
            inner join cv on cv.cv_id = person.person_cv";

// EXECUTE QUERY IF THE RESULT BIGGER IS THAN ZERO
// VOER QUERY UIT ALS RESULTAAT GROTER IS DAN NUL
$result = $conn->query($sql);
if ($result->num_rows > 0) {

// ECHO A TABLE WITH VALUES 
// ECHO EEN TABEL MET ONDERSTAANDE WAARDES
echo "<form action='admin.php' method='post'>"; 
echo "<table border=0 align=right>
<tr>
<th>Voornaam</th>
<th>Achternaam</th>
<th>Straat</th>
<th>Huisnummer</th>
<th>Postcode</th>
<th>Stad</th>
<th>Provincie</th>
<th>Email</th>
<th>Mobiel</th>
<th>cv</th>
<th>delete</th>
</tr>";

// LOOP THROUGH THE RESULTS AND OUTPUT THE RESULTS FOR EACH ROW
// GA DOOR RESULTATEN EN LAAT DE RESULTEN PER RIJ ZIEN
while($row = $result->fetch_assoc()) {
     echo "<tr>";
     echo "<td>" . $row["person_firstname"] . "</td>";
     echo "<td>" . $row["person_lastname"] . "</td>";
     echo "<td>" . $row["address_street"] . "</td>";
     echo "<td>" . $row["address_housenumber"] . "</td>";
     echo "<td>" . $row["address_zipcode"] . "</td>";
     echo "<td>" . $row["address_city"] . "</td>";
     echo "<td>" . $row["address_state"] . "</td>";
     echo "<td>" . $row["person_email"] . "</td>";
     echo "<td>" . $row["person_phonenumber"] . "</td>";
     echo "<td><a href='http://localhost:8080/website/" . $row['cv_path'] . "'>cv file</a></td>";
     echo "<td><a href='delete.php?person_id=" . $row['person_id'] . "'>delete</a></td>";
     echo "</tr>";
  }
  echo "</table>";
  echo "</form>";
 }

  // IF THERE IS ZERO RESULT ECHO THIS
  // ALS WAARDE NUL IS LAAT DE ONDERSTAADE TEKST ZIEN
  else {
    echo "<p id='Tekst'>Er zijn geen deelnemers in de database gevonden.</p>";
  }
  // CLOSE CONNECTION
  // SLUIT CONNECTIE
  $conn->close();
 ?>

Answer 1

您只需将默认约束设置为＆＃34; No Resume＆＃34;。

创建表时，只需确保cv的默认值设置为＆＃34;没有恢复＆＃34;，当您需要调用详细信息时，只需检查该字段是否有＆＃34;没有简历＆＃34;价值，如果它不是，只是这个 -

$_POST

Answer 2

在里面

你这样做

 echo "<td><a href='http://localhost:8080/website/" . $row['cv_path'] . "'>cv file</a></td>";

试试这个

$path = "http://localhost:8080/website/" . $row['cv_path'];
if(!empty($row['cv_path']) && file_exists($path)){
     echo "<td><a href='$path'>cv file</a></td>";
}
else{
     echo "<td>No Resume</td>";
}

链接为空时如何显示消息

2 个答案: