PHP的查詢，並保持在選定

Question

我想MySQL查詢，並保持在選擇下拉列表

<select name="selectjob"></select> 

，這是MySQL查詢

SELECT repair_id FROM `repair_list` where status = 'work' 

Answer 1

所有你需要做的是：

//db connection 
    mysql_connect("localhost","user","password"); 
    mysql_select_db("database"); 

    //query 
    $sql=mysql_query("SELECT repair_id FROM repair_list where status = 'work'"); 
    if(mysql_num_rows($sql)){ 
    $select= '<select name="selectjob">'; 
    while($rs=mysql_fetch_array($sql)){ 
      $select.='<option value="'.$rs['repair_id'].'">'.$rs['repair_id'].'</option>'; 
     } 
    } 
    $select.='</select>'; 
    echo $select; 

或者使用mysqli作爲我的天才朋友在建議的評論中：

<?php 

$conn = new mysqli('localhost', 'username', 'password', 'database') 
or die ('Cannot connect to db'); 

    $result = $conn->query("SELECT repair_id FROM repair_list where status = 'work'"); 

    echo "<select name='selectjob'>"; 
    while ($row = $result->fetch_assoc()) { 
     echo '<option value="'.$row['repair_id'].'">'.$row['repair_id'].'</option>'; 
    } 
    echo "</select>"; 
?> 

Answer 2

這將引導您完成：

$status = 'work'; 

$query = $mysqli->prepare("SELECT repair_id FROM `repair_list` where status = ?"); 

$query->bind_param("s",$status); 

if (!$query->execute()) 
{ 
    $flag = false; 
} 

$query->bind_result($repair_id); 

$select= '<select name="selectjob">'; 
while($query->fetch()){ 
     $select.='<option value="'.$repair_id.'">'.$repair_id.'</option>'; 
    } 
} 
$select.='</select>'; 

Answer 3

select.php

$stmt = $dbh->prepare("SELECT repair_id FROM repair_list where status ='work'"); 
$selected_row = $stmt->fetch(PDO::FETCH_ASSOC); 
$stmt->execute(); 
if ($stmt->rowCount() > 0) { 
     while ($selected_row = $stmt->fetch(PDO::FETCH_ASSOC)) { 
      $repair_id[] = array('repair_id' => $selected_row['repair_id ']); 
     } 
     echo json_encode($repair_id, JSON_UNESCAPED_UNICODE); 
    } 

JS

$.ajax({ 
    type: 'POST', 
    url: 'select.php', 
    dataType: "json", 
    success: function(data) { 
     var select = ('#select'); 
      for (var i = 0; i < data.length; i++) { 
       $(select).append('<option id=' + data[i].repair_id + ' value="' + data[i].repair_id + '">' + data[i].repair_id + '</option>'); 

      } 
    }, 
    error: function(data) { 
    } 
}); 

html

<select name="selectjob" id="select"></select> 

嘗試使用AJAX

Answer 4

$servername = "localhost"; 
$username = "username"; 
$password = "password"; 
$dbname = "myDB"; 


$conn = new mysqli($servername, $username, $password, $dbname); 

if ($conn->connect_error) { 
    die("Connection failed: " . $conn->connect_error); 
} 

$sql = "SELECT repair_id FROM repair_list where status = 'work'"; 
$result = $conn->query($sql); 

if ($result->num_rows > 0): 
    $html =''; 
    while($row = $result->fetch_assoc()) { 
     $html.='<option value="'.$row['repair_id'].'">'.$row['repair_id'].'</option>'; 
    } 
    echo $html; 
else: 
    echo "0 results"; 
endif; 
$conn->close(); 

Answer 5

你是不是清楚，您保存所選repair_id。通過更改$ selectedValue變量的值，您可以顯示所需的選定值。

<select name="selectjob"></select> 

    <?php 
    $query = mysql_query("SELECT repair_id FROM `repair_list` where status = 'work'"); 
    $selectedValue = 2; //some id which is selected 
    while ($row = mysql_fetch_array($query)) { ?> 
    <option value="<?php echo $row['repair_id']?>" <?php echo ($row['repair_id'] == $selectedValue)? ' selected="selected"': ''?>><?php echo $row['repair_id'];?></option> 
    <?php } ?>