我有一个保存在 MYSQL 数据库中的画布图像,可以在使用数据表和 PHP 的表中看到,但我无法下载图像。这是我的 JS 文件,它将 ajax 请求发送到服务器:$(document).ready(function(){ var data = $('#dataList').DataTable({ "lengthChange": false, "processing":true, "order":[], "ajax":{ url:"/php/process.php", type:"POST", data:{action:'listData'}, dataType:"json" }, "columnDefs":[ { "targets":[0, 5, 6], "orderable":true, }, ], "pageLength": 10 });这是process.php$sqlQuery = "SELECT * FROM table1 AS a LEFT JOIN sketch AS s ON a.id= s.id";$auftragData = array();$result = $this->dbc -> prepare($sqlQuery);$result -> execute(); while ( $tableResult= $result->fetch(PDO::FETCH_ASSOC) ) { $resultRows = array(); $resultRows[] = $tableResult['id']; $resultRows[] = ucfirst($tableResult['cust_id']); $resultRows[] = $tableResult['typ']; $resultRows[] = $tableResult['status']; $resultRows[] = $tableResult['sketch']; if ($tableResult['sketch']) { $resultRows[] = '<a id="download" download="sketch.png"><button type="button">Download Image</button></a>'; } $resultRows[] = '<button type="button" name="update" id="'.$tableResult["id"].'" class="btn btn-warning btn-xs update">update</button>'; $resultRows[] = '<button type="button" name="delete" id="'.$tableResult["id"].'" class="btn btn-danger btn-xs delete" >delete</button>'; $finalData[] = $auftragRows; }$numRows = $result -> rowCount(); $output = array( "draw" => $numRows, "recordsTotal" => $numRows, "recordsFiltered" => $numRows, "data" => $finalData ); echo json_encode($output); $this->dbc = NULL; exit; }该图网址为:数据:图像/ PNG; BASE64,iVBORw0KGgoAAAANSUhEUgAAA4QAAAH0CAYAAABl8 + PTAAAgAElEQVR4Xu3dB9glZ103 / u8LJCHhpQkYCUWlWwAFFZDQO0F6lUAkFEEpghQpvuArRboUAaU36YYWpBMhFP1Lk1clIhZAOoIgCUko / + UHS / pks7vPmTkzZ2bOfOa6zrULue如何使用上述代码下载图像?
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