0

您好，我一直在试图弄清楚为什么我的旨在在表中列出查询结果的代码不起作用。我采用了在网上找到的代码，并且厌倦了适应它。我的数据存储在 pgsql 中。html 页面有一个允许选择机构名称的下拉菜单。当我单击提交按钮以了解我的数据库中谁属于该机构时，php 页面已加载并显示我要发送到 pgsql 的 SQL 查询。我应该得到显示在 html 页面上的表格中的查询结果。下拉菜单正常工作，因此我不提供此菜单的 php 代码（listinstitutions.php）

有人告诉我应该使用 ajaxsubmit() 但我不知道该函数放在哪里。php 文件中也可能有错误，因为查询被显示而不是被发送到 pgsql。json 是否正确发送？

您的指导将不胜感激。

谢谢你。

html端：

<html>
<head>

<link rel="stylesheet" href="http://code.jquery.com/ui/1.10.1/themes/base/jquery-ui.css" />

<script src="http://code.jquery.com/jquery-1.9.1.min.js"></script>
<script src="http://code.jquery.com/ui/1.10.1/jquery-ui.js"></script>
<script>
$(document).ready(function(){


  //////////////////////////////////////
  // Displays insitution names in Drop Down Menu

        //Getting the selector and running the code when clicked
        $('#selinstit').click(function(e){
                 //Getting the JSON object, after it arrives the code inside
               //function(dataI) is run, dataI is the recieved object
               $.getJSON('http://localhost/listinstitutions.php',function(dataI){
                            //loop row by row in the json, key is an index and val the row
                            var items = [];  //array
                          $.each(dataI, function(key, val) {
                            //add institution name to <option>
                            items.push('<option>' + val['Iname'] + '</option>');
                        });//end each
                        //concatenate all html
                        htmlStr=items.join('');
                        console.log(htmlStr);
                        //append code
                        $('option#in').after(htmlStr);
                });//end getJSON
        });//end cluck



    ///////////////////////////////
   // Displays persons form an institution in a table

     $( "$subinst" ).button().click(function( event ) {
     console.log($(this)); // for Firebug
     $.getJSON('http://localhost/SelectPersonsBasedOnInstitution.php',function(data){   // I make an AJAX call here
     console.log($(this)[0].url); // for Firebug   check what url I get here
                            //loop row by row in the json, key is an index and val the row
                            var items = [];  //array
                          $.each(data, function(key, val) {

                        //add table rows
                            items.push('<tr border=1><td>' + val['Pfirstname'] + '</td><td>' + val['Plastname'] + '</td><td><a mailto:=" ' + val['Pemail'] + ' " >' + val['Pemail'] + '</a></td></tr>');
                        });//end each
                        //concatenate all html
                    htmlStr=items.join('');

                        //append code
                        $('#instito').after(htmlStr);
                });//end getJSON
      event.preventDefault();
      });

}); //end ready

</script>

</head>   

<body>
<form id="myForm" action="SelectPersonsBasedOnInstitution.php" method="post">
Select persons from an institution:
<br>                                            
<tr>
 <td>
   <select id="selinstit" name="instit">
   <option id="in">Select</option>                       
   </select>
 </td>
 <td>
   <input type="submit" id="subinst" value="Submit" /> 
 </td>
</tr>

</form>

   <table frame="border" id="instito">
   </table>
</body>
</html>

这是 SelectPersonsBasedOnInstitution.php 的 php 代码

<?php


//////////
// part 1: get information from the html form
ini_set('display_errors', 1);                                      
ini_set('display_startup_errors', 1);

foreach ($_REQUEST as $key => $value){
 $$key=$value;  
}

// part2: prepare SQL query from input
$sqlquery= sprintf('SELECT "Pfirstname", "Plastname", "Pemail" FROM "PERSON"
LEFT JOIN "INSTITUTION" ON
"PERSON"."Pinstitution"="INSTITUTION"."Iinstitution"
WHERE "Iname" = \'%s\'',$instit);
echo $sqlquery;


/////////
// part3: send query
$dbh = pg_connect("host=localhost dbname=mydb user=**** password=*****");
$sql=  $sqlquery;
$result = pg_query($dbh,$sql);
$myarray = pg_fetch_all($result);

$jsontext = json_encode($myarray);
echo($jsontext);

?>
4

1 回答 1

1

以下行可能是问题所在（不应该存在）：

echo $sqlquery;

在没有那行的情况下重写你的代码，它应该可以工作。

$sqlquery= sprintf('SELECT "Pfirstname", "Plastname", "Pemail" FROM "PERSON" LEFT JOIN "INSTITUTION" ON "PERSON"."Pinstitution"="INSTITUTION"."Iinstitution" WHERE "Iname" = \'%s\'', $instit);

$dbh = pg_connect("host=localhost dbname=mydb user=**** password=*****");
$result = pg_query($dbh, $sqlquery);
$myarray = pg_fetch_all($result);
$jsontext = json_encode($myarray);
echo($jsontext);
于 2013-03-27T17:11:41.003 回答