JQuery AJAX請求的答案，但沒有成功

我有這個文本框用於檢查數據庫上的名稱，它提供了答案和一切，但如果我把console.log(data)它不回答。JQuery AJAX請求的答案，但沒有成功

這裏是我的代碼：

  function absurdSearch(textoBuscar){ 
      $.ajax({ 
      type : "POST", 
      url  : "service.php", 
      dataType: "json", 
      data : { 
       action:"absurdSearch", 
       absurdText: textoBuscar 
       },   
       success:function(data){ 
        console.log(data); 
        $("#fotoproyectosarq").empty(); 
        var html = ''; 
        html += '<img src="' + data.path + '" height="128" width="160">'; 
        $("#fotoproyectosarq").append(html); 

        $("#nombreproyectosarq").empty(); 
        var html2 = ''; 
        html2 += "<form method=\"post\" name=\"projectsearch\" id=\"projectsearcharq\" action=\"proyectos_arq.php\">" 
        html2 += "<span style=\"cursor: pointer;\" onclick=\"document.getElementById('projectsearcharq').submit()\">"+ data.projectName +"</span>" 
        //html2 += "<button id=" + "button" + data.projectId + " style=\"visibility:hidden;\"><span id=" + data.projectId + " style=\"cursor: pointer;\"><span>" + data.projectName + "</span></span></button>" 
        html2 += "<input name=\"project_id\" type=\"hidden\" id=\"project_id\" value=" + data.projectId + ">" 
        html2 += "</form>" 
        $("#nombreproyectosarq").append(html2); 


       } 
      }) 
     } 

    $('#buscadorRapidoTextInput').on('keyup', function() { 
     var textoBuscar = $(this).attr('value'); 
     absurdSearch(textoBuscar); 
    });

和服務answersaccording到螢火蟲：

memphis{"projectCategory":"1","projectId":"5","projectName":"MEMPHIS RIVERFRONT","path":"server\/php\/files\/MF_1.jpg"}

但成功事件似乎並沒有工作。

這裏的後端代碼：

function absurdSearch($absurdText){ 

    $db = new db(); 
    $conn = $db->conn(); 

    $SQL_ABSURD_SEARCH="SELECT projects.id as projectId, projects.project_types_id as projectCategory, projects.name as projectName, projects.description as description, path FROM projects INNER JOIN project_types ON (projects.project_types_id = project_types.id) inner join images on images.projects_id = projects.id where images.main = '1' and projects.name LIKE '%$absurdText%' LIMIT 1"; 
    $conn->query($SQL_ABSURD_SEARCH); 
      foreach($conn->query($SQL_ABSURD_SEARCH) as $row) { 
       $projectCategory = $row['projectCategory']; 
       $projectId = $row['projectId']; 
       $projectName = $row['projectName']; 
       $path = "server/php/files/".$row['path'].""; 

       $rows = array("projectCategory" => $projectCategory,"projectId" => $projectId, "projectName" => $projectName, "path" => $path);    

       $json = json_encode($rows); 

       echo $json; 
      } 
}

來源

2012-12-19 Solar Confinement

也許我錯了，但是這是每一個產生太多的html按鍵 – Ibu

我認爲這只是異步問題。嘗試在選項中將'async'設置爲'false'。 –

嘗試一個警報，這將告訴你，如果它是成功的。 –

需要調用

exit(json_encode($rows);

這應該解決您的問題

來源

2012-12-20 04:39:48

JQuery AJAX請求的答案，但沒有成功

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