检索单个列的值

Question

我需要从单个列中检索数据并将其放入API（Json）中，但出于某种原因，我也从列中获取标题。

$sql = "SELECT workingJson FROM dataTable"; 

我认为它会像workingJson.Value，但没有运气。

这里是API.php根据您的意见

// Create connection 
$con=mysqli_connect("localhost","user","password","database"); 

// Check connection 
if (mysqli_connect_errno()) 
{ 
    echo "Failed to connect to MySQL: " . mysqli_connect_error(); 
} 

// This SQL statement selects ALL from the table 'Locations' 
$sql = "SELECT column1 FROM database"; 

// Check if there are results 
if ($result = mysqli_query($con, $sql)) 
{ 
    // If so, then create a results array and a temporary one 
    // to hold the data 
    $resultArray = array(); 
    $tempArray = array(); 

    // Loop through each row in the result set 
    while($row = $result->fetch_object()) 
    { 
     // Add each row into our results array 
     $tempArray = $row; 
     array_push($resultArray, $tempArray); 
    } 

    // Finally, encode the array to JSON and output the results 
    echo json_encode($resultArray); 
} 

// Close connections 
mysqli_close($con); 

Answer 1

编辑：

要只返回在PHP中的值，而不是钥匙，你可以在你的代码改成这样：

// Loop through each row in the result set 
while($row = $result->fetch_object()) 
{ 
    // Add value to array 
    array_push($resultArray, $row->column1); 
} 

在这种情况下，您不需要$tempArray。

Answer 2

你可以得到所有的结果和做traitment后：

<?php 

// Create connection 
$con=mysqli_connect("localhost","user","password","database"); 

// Check connection 
if (mysqli_connect_errno()) 
{ 
    echo "Failed to connect to MySQL: " . mysqli_connect_error(); 
} 

// This SQL statement selects ALL from the table 'Locations' 
$sql = "SELECT column1 FROM database"; 


if ($result = $mysqli->query($sql)) { 

    //get all fields : 
    $finfo = $result->fetch_fields(); 

    foreach ($finfo as $val) { 
     $resultArray[] = $val->column1; 
    } 
    $result->close(); 
} 

// Finally, output the results without encode because is also in json format: 
echo '{'. implode(','.chr(10), $resultArray) .'}'; 

$mysqli->close();