Question

您好我一直在试图找出为什么我的代码旨在列出表中的查询结果不起作用。我在网上找到了代码并厌倦了调整它。我的数据存储在pgsql中。 html页面有一个下拉菜单，允许选择机构名称。当我单击提交按钮以了解我的数据库中属于该机构的人时，将加载php页面并显示我要发送到pgsql的SQL查询。我应该在html页面上显示的表格中显示查询结果。下拉菜单正常工作，所以我不提供这个的代码（listinstitutions.php）

有人告诉我应该使用ajaxsubmit（），但我不知道在哪里放这个功能。由于显示查询而不是发送到pgsql，因此php文件中可能还有错误。 json是否正确发送了？

非常感谢您的指导。

谢谢。

html方面：

<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);

?>

Answer 1

以下行可能是问题（它不应该存在）：

echo $sqlquery;

重写没有该行的代码，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);

使用AJAX显示sql查询而不是查询结果

1 个答案: