Question

我有一个依赖于上一页的php页面。根据我点击的客户类型，我将在下一页上看到他们的信息。我现在有一个问题，我必须将图形嵌入到php页面中，以便为该客户检索SQL数据。因为如果我在单独的PHP页面上有图表，它不知道我点击哪个客户来显示他们的统计数据。

如何将jpgraph嵌入其他php？

PHP

<?php

// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

$result = mysqli_query($mysqli,"SELECT rid, pid, firstname, lastname, email FROM temp_members_db WHERE pid='$pid1' ");

echo "<table><tr><th>Namn</th><th>E-mail</th><th>Resultat</th><th>Ta bort kandidat</th></tr>";
while($row = mysqli_fetch_array($result))
{

$count= "SELECT COUNT(qid) AS 'Total' FROM result WHERE rid=".$row['rid']."";
$result01 = mysqli_query($mysqli, $count); 
$resultArr01 = mysqli_fetch_array($result01);
$total= $resultArr01[0]*10;

$pointsummary= "SELECT SUM(points) AS 'Summary' FROM result WHERE rid=".$row['rid']."";
$result02 = mysqli_query($mysqli, $pointsummary);
$resultArr02 = mysqli_fetch_array($result02);

$sum= $resultArr02[0];

$result00 = ($total != 0 ? round($sum/$total*100) : 0);
  
	

    
    $color = "#000000";

if (($result00 >= 1) && ($result00 <= 30))
   $color = "#FF0000";
else if (($result00 > 30) && ($result00 <= 60))
   $color = "#FF9900";
else if (($result00 > 60) && ($result00 <= 100))
   $color = "#07d407";

echo "<tr><td><strong>
<form action='/devlopment/graph.php' method='GET'>
 <input type='hidden' name='rid' value='".$row['rid']."'>
 <input type='hidden' name='firstname' value='".$row['firstname']."'> 
<input type='submit' class='resname' name='submit' value='".$row['firstname']." ".$row['lastname']."'>

 </form>
 </strong></td> 
 <td>".$row['email']."</td> 
 <td><strong><span style=\"color: $color\">".$result00."</span>%</strong></td>   
 
 <form action='deleterespondent2.php' method='post'>
 <input type='hidden' name='rid' value='".$row['rid']."'> 
<td> <input type='submit' class='mydel' value='Radera' onclick=\"return confirm('Vill du verkligen radera kandidaten? \\n\\nObservera att kandidatens eventuellt besvarade frågor och resultat raderas också!')\">
 </form>
</td></tr>";

}
echo "</table>";
mysqli_close($mysqli);
?>

图表

<?php 
require_once('jpgraph/jpgraph.php');  
require_once('jpgraph/jpgraph_line.php');  
require_once('jpgraph/jpgraph_bar.php'); 
session_start();
$val = (isset($_GET['rid']))  ? 
    "{$_GET['rid']}" : '';  
 
$x_axis = array();  
$y_axis = array();  
$i      = 0;  
$con    = mysqli_connect("root", "g", "f", "d");  
// Check connection  
if (mysqli_connect_errno()) {  
    echo "Failed to connect to MySQL: " . mysqli_connect_error();  
}  
$result = mysqli_query($con, "SELECT * FROM result WHERE rid=$val");  
while ($row = mysqli_fetch_array($result)) {  
    $x_axis[$i] = $row["qid"];
    $y_axis[$i] = $row["points"];
    $i++;  
}  


mysqli_close($con);  
$width  = 600;  
$height = 200;  
$graph  = new Graph($width, $height);  
$graph->SetScale('textint');  
$graph->title->Set('My Example');  
$graph->xaxis->title->Set('(Fråga)');  
$graph->xaxis->SetTickLabels($x_axis);  
$graph->yaxis->title->Set('(Poäng)');  
$barplot = new BarPlot($y_axis);  
$graph->Add($barplot);  
$graph->Stroke();
?>

Answer 1

这可能要晚了，但是为了以防万一或对于这里的其他人，您真的不需要将JGraph放在同一页上。但是，嵌入将是正确的。基于以上所述，您可以执行以下操作：

//graphshow.php
<?php 
require_once('jpgraph/jpgraph.php');  
require_once('jpgraph/jpgraph_line.php');  
require_once('jpgraph/jpgraph_bar.php');

if (isset($_GET['axes'])) {
    $axes = explode("^^", $_GET['axes']);
    $x_axis = $axes[0];
    $y_axis = $axes[1];
}

$width = 600;  
$height = 200;  
$graph = new Graph($width, $height);  
$graph->SetScale('textint');  
$graph->title->Set('My Example');  
$graph->xaxis->title->Set('(Fråga)');  
$graph->xaxis->SetTickLabels($x_axis);  
$graph->yaxis->title->Set('(Poäng)');  
$barplot = new BarPlot($y_axis);  
$graph->Add($barplot);  
$graph->Stroke();
?>

然后在您的主php中，您需要在其中显示图形：

<?php
session_start();
$val = (isset($_GET['rid'])) ? "{$_GET['rid']}" : '';  
$x_axis = array();  
$y_axis = array();  
$i = 0;  
$con = mysqli_connect("root", "g", "f", "d");  
// Check connection  
if (mysqli_connect_errno()) {  
    echo "Failed to connect to MySQL: " . mysqli_connect_error();  
}  
$result = mysqli_query($con, "SELECT * FROM result WHERE rid=$val");  

while ($row = mysqli_fetch_array($result)) {  
    $x_axis[$i] = $row["qid"];
    $y_axis[$i] = $row["points"];
    $i++;  
}
$xy = $x_axis."^^".y_axis; 
?>

<div>
    <embed src="graphshow.php?axes=<?php echo $xy; ?>" height=80em; width=160em;> </embed>
</div>

嵌入jpgraph与其他PHP

1 个答案: